"We must see that the clock is working as normal although when we move away from the clock at the speed of light." Elbart Einstine said if we go away from a clock( clock tower) at the speed of light, we will see the clock is freezed. But If the speed of light is constant and non relative. So When we go away from a clock at the speed of light, we can't see that light coming from clock is stopped because the speed of light doesn't depend on our speed ( speed of light is not relative). So even when we go at the speed of light. We must see the light of the clock tower going at the speed of light. So how do we observe that the clock of the tower is freezed. I don't think it is freezed. "We must see that the clock is working as normal although when we move away from the clock at the speed of light." Right?
@@MichelvanBiezen yeah. It was said by elbart. But nobody saw it. Anyway my key point of this problem is this, if we say that we see the clock is freezed, we are considering light under dopler effect. But the dopler effect happens in sound ( sound is relative) Do we use same doplar effect theory to understand the scenario of person going away from clock at light speed
We do see the doppler effect when light emanates from a moving source. This however is not the doppler effect. It is the relativistic effect that occurs when the relative speeds approach the speed of light. This has been confirmed numerous times in the laboratory and by using accelerators.
I found this video by typing "relativistic tidal effects." I was looking the relativistic effects caused by gravity. Especially by filaments. I'm wondering if this can account for so called dark energy. Perhaps filaments are three or four orders of magnitudes more massive than measurements suggest. Can the tidal effect between such massive objects cause relativistic tidal effects known as "dark energy?"
Very nice presentation. Thanks. I would also recommend to put the classical form of Doppler next to it and compare . That's very interesting, because for typically measurable Doppler effect, the relativistic formula gives a quite different frequency than that of classic one. So if this measured in reality (which is certainly so) , it is another proof for relativistic formula.
im confusing some part of this. when you write f=c/((c-u)T) at around 2:57 , isnt f defined as 1/T and therefore the substitution could be made (1/T)=c/((c-u)T) ? this doesnt make sense to me please help me understand what im doing wrong.
and by that same argument if T=gamma*T_0, finding the reciprocal of both sides, f = f_0*(1/gamma). i know this isnt right but i dont quite understand why.
There are a lot substitutions you could make with physics equations and mathematical methods, but it doesn't mean they will be helpful to your objective. He is trying to show how to find the frequency when taking in account the difference in time as observed by the stationary observer. To do this, he uses the equation as described @3:15. I think you could make the substitution you made, but you would still substitute T on the right side of your equation for T_0/(sqrt(1 - u2/c2)). Then, in order to solve for T, you would have to get the multiplicative inverse of everything on the right side of the equation. It's just a little more work to make the substitution you made and find the period instead of the frequency, and who wants to do more work?
4 года назад
@@Pete133 The question is not to do more or less work, but to do a correct work. - by definition, f=1/T and fo=1/To (cf [4:40] - at [3:03] he writes f=c/((c-u)T). This would be correct only if u=0 That means this derivation does not work, sorry.
Noé Neuville - My point was that I would rather not do extra work by making unnecessary substitutions if it doesn’t help with the objective. I don’t understand why you are positing that f=c/((c-u)T) is only correct when u=0. Care to explain?
@ Also a fair warning... so you don't waste your time... if you have to disprove Special Relativity to make your point, it's probably going to be over my head. I would watch some of your videos, but I don't understand French. Either way, the fact that Richard Feynman taught the same relativistic doppler effect ideas makes me think that I'm not likely to disprove or even understand a disproving of it.
1:20 But if the speed of light is always c, then how does the source "catch up" to the first wave crest emitted? How can the wavelength change if light is always moving away from the source with c no matter its own velocity?
If an object moves away from you at velocity v1 and you walk in the same direction at velocity v2, then you will be closer to the object at any time then if you had not moved at all.
@@MichelvanBiezen But if that's true, wouldn't an observer at the source measure a speed of light slower than c? Namely c-v or c+v in the other direction (v=sources velocity)
If relativistic objects emit radiation focused in the forward direction ("headlight effect"), then do they reflect radiation the same way? If so, then if any relativistic object reflects starlight, radar, etc., then it only does so towards its forward direction of travel, and is nearly invisible to the sides or rear?
The first principle that Einstein used in his formulation of the special theory of relativity is that every observer, regardless of how fast the observer is traveling will see light (E&M radiation) approach and move away at the speed of light.
@@MichelvanBiezen Relativistic particles, which radiate isotropically in their own rest frames, appear to radiate preferentially in the forward direction to others? So, if there was some relativistic object hurtling through our Solar System, reflecting visible-wavelength ambient sun- and star-light, then that object would appear to be bright in gamma rays on the way in, when earth observers were within the "headlight angle" of order 1/gamma. As the object began to pass, it would dim, and the radiation redshift, into x-rays & UV, then dim & redshift further into visible as it passed by at a LOS 90 degrees to its travel. As began to whoosh away, its apparent radiation would dim & redshift further into IR, millimeter, microwave & fade away into the distance as an ultra faint radio source. To detect such an object would require some sort of multi-band, omni-band, all-band, full-spectrum, gamma-to-radio detector capable of imaging everything from ultra bright to ultra faint. If a human observer were in orbit, aboard the ISS, unable to see anything but visible wavelengths, they would only sense a brief, fading, visible flash of a streak as the object whooshed past the 90-degree point of closest approach to them. They would conclude cosmic rays were affecting their eyes, and think nothing further of it. Not sure how you could build an all-spectrum detector, except perhaps to just attach other kinds of detectors (visible, IR, UV) into the gaps in the framework of a large radio dish. From what I understand right now, something like that would be necessary to recognize the passage of a near-lightspeed object. One would have to overlay all the images from all the different frequency bands in order to recognize that they all "stitched together" into one long streak of a gamma-bright-to-radio-dim track (rather than being a bunch of independent flashes in each of the bands).
Math Geeks Since we do not care about the coordinates, we can just assume that x in the equation that you have written is 0, and therefore T=gamma•T_0. There is also another simple method to prove the time dilation equation without any coordinate-system. And the result is that T=gamma•T_0.
Hello sir, how did you get from c-u to radical and squared that?it is at 6.49. I didn't get the transition. Did you multiply or how? Would be grateful if you reply and thanks a lot for your videos!
Very good question. When v = 10 % of c (the speed of light), then your error will be 0.5 % if you don't use the Lorentz transformation equations. When v = 50 % of c, the error will be 15% and if v = 86.7 % of c then your error will be 100%..
I'm not that far along in physics just yet, but does this mean the speed at which I am moving away from the source helps to determine the amount of red shift observed?
Thanx for your videos :) I'm more a philosopher than a physician but anyway, let's imagine two cases: . 1. A star is moving away very fast from an observer on the earth, red-shifted light with a long wave length. . 2. A star is approaching very fast to an observer on the earth, blue-shifted light with a short wave length. . Let's assume that the two stars are now at exactly the same distance from earth. . Then let's, theroretically, draw these two waves as two separate straight lines instead (stretch them out). The blue line (former wave) will then be longer than the red one. Is it possible to explain, with this little thought experiment, that the speed of light emanating from the two objects is the same for both of them (independent of moving away or approaching) ? Basically that photons have to travel a longer distance ("riding on or within" the blue wave), so that the speed in a straight direction (speed of light) stays the same as with the red wave ? . Trying to get a grip of this in a real/deeper physical sense if possible, why the speed of light stays the same when objects move differently relative to an observer. Is that a phenomenon which is dependent on some rate of change within the object itself when moving (compared to a rate of change in the observer as an object), or can it be explained occuring within the communication between the objects (my example) ?
Question sir michel. What if the observer also travels nearly as fast as the source of light, will the Classical Doppler's Effect be used instead of Relativistic Doppler's Effect? Sorry for this question sir.
If the observer travels near the speed of light, all the equations of relativity play a role and yes, the observer will still see light moving at the speed of light regardless of what the source is doing.
@@MichelvanBiezen so how would you relate the frequencies measured by source and observer in that case? t = tO/sqrt(1 + v^2/c^2). tO is applied when the observer isnt moving but it that case, it is moving. Will it be tS = tO/sqrt(1 + vS^2/c^2) and tA = tO/sqrt(1 + vA^/c^2) Where tS = time measured by source. tA = time measured by observer. t = time measured by stationary object. vS = speed of source moving nearly at c. vA = speed of observer moving nearly at c. If not, pls make vid about this for clarifications. Thanks sir for your reply.
Question sir. What if the obsever also moving nearly at speed of light? Will he/she measure "c" (based on 2nd postulate of special relativity) or will he/she measure "c +or- v" where "v" is his/her speed nearly speed of light depending if he/she is away or towards the source?
well... im 5 months late if the observing and the source of light are moving towards each other with velocities u and v respectively then from the reference frame of the observer .. he is at rest but now he will see the light source moving towards him with a velocity of v+u
Have to say, for a college student in undergraduate physics classes, your videos are so useful! Thanks and keep up the great work!
Wonder if u finished ur degree :D
Probably the one of the best teacher..a thousand time better way of explanation than my college teachers. Love your way of teaching...thank you
please make a video on relativistic Doppler effect when observing under an angle
you are good at explaining physics and helping out people.
Wow - I wasn't expecting to be able to follow, but it actually made sense... thank you Michel van Biezen!
perfectly explained. Thank you very much.
I'm privileged always i watch your video sir, i only heard of Einstein but you're the one for me.
Thank you. Glad you are enjoying our videos. 🙂
Thank you sir..
Your way of explaining makes it easy to understand better. I've understood it very well
"We must see that the clock is working as normal although when we move away from the clock at the speed of light."
Elbart Einstine said if we go away from a clock( clock tower) at the speed of light, we will see the clock is freezed.
But If the speed of light is constant and non relative.
So
When we go away from a clock at the speed of light, we can't see that light coming from clock is stopped because the speed of light doesn't depend on our speed ( speed of light is not relative). So even when we go at the speed of light. We must see the light of the clock tower going at the speed of light.
So how do we observe that the clock of the tower is freezed.
I don't think it is freezed.
"We must see that the clock is working as normal although when we move away from the clock at the speed of light."
Right?
Einstein is correct. If you approach, (or move away from) a clock at the speed of light, the clock will appear to stand still.
@@MichelvanBiezen yeah. It was said by elbart. But nobody saw it. Anyway my key point of this problem is this, if we say that we see the clock is freezed, we are considering light under dopler effect. But the dopler effect happens in sound ( sound is relative)
Do we use same doplar effect theory to understand the scenario of person going away from clock at light speed
We do see the doppler effect when light emanates from a moving source. This however is not the doppler effect. It is the relativistic effect that occurs when the relative speeds approach the speed of light. This has been confirmed numerous times in the laboratory and by using accelerators.
@@MichelvanBiezen thank you
it was nice video. for physics student like me, this video is very useful. thanks and keep it up!
I found this video by typing "relativistic tidal effects." I was looking the relativistic effects caused by gravity. Especially by filaments. I'm wondering if this can account for so called dark energy. Perhaps filaments are three or four orders of magnitudes more massive than measurements suggest. Can the tidal effect between such massive objects cause relativistic tidal effects known as "dark energy?"
Very nice presentation. Thanks.
I would also recommend to put the classical form of Doppler next to it and compare .
That's very interesting, because for typically measurable Doppler effect, the relativistic formula gives a quite different frequency than that of classic one. So if this measured in reality (which is certainly so) , it is another proof for relativistic formula.
im confusing some part of this. when you write f=c/((c-u)T) at around 2:57 , isnt f defined as 1/T and therefore the substitution could be made (1/T)=c/((c-u)T) ? this doesnt make sense to me please help me understand what im doing wrong.
and by that same argument if T=gamma*T_0, finding the reciprocal of both sides, f = f_0*(1/gamma). i know this isnt right but i dont quite understand why.
There are a lot substitutions you could make with physics equations and mathematical methods, but it doesn't mean they will be helpful to your objective. He is trying to show how to find the frequency when taking in account the difference in time as observed by the stationary observer. To do this, he uses the equation as described @3:15. I think you could make the substitution you made, but you would still substitute T on the right side of your equation for T_0/(sqrt(1 - u2/c2)). Then, in order to solve for T, you would have to get the multiplicative inverse of everything on the right side of the equation. It's just a little more work to make the substitution you made and find the period instead of the frequency, and who wants to do more work?
@@Pete133 The question is not to do more or less work, but to do a correct work.
- by definition, f=1/T and fo=1/To (cf [4:40]
- at [3:03] he writes f=c/((c-u)T). This would be correct only if u=0
That means this derivation does not work, sorry.
Noé Neuville - My point was that I would rather not do extra work by making unnecessary substitutions if it doesn’t help with the objective.
I don’t understand why you are positing that f=c/((c-u)T) is only correct when u=0. Care to explain?
@ Also a fair warning... so you don't waste your time... if you have to disprove Special Relativity to make your point, it's probably going to be over my head. I would watch some of your videos, but I don't understand French. Either way, the fact that Richard Feynman taught the same relativistic doppler effect ideas makes me think that I'm not likely to disprove or even understand a disproving of it.
1:20
But if the speed of light is always c, then how does the source "catch up" to the first wave crest emitted?
How can the wavelength change if light is always moving away from the source with c no matter its own velocity?
If an object moves away from you at velocity v1 and you walk in the same direction at velocity v2, then you will be closer to the object at any time then if you had not moved at all.
@@MichelvanBiezen But if that's true, wouldn't an observer at the source measure a speed of light slower than c? Namely c-v or c+v in the other direction (v=sources velocity)
The speed of light is not changing, but the crests of light will be bunched up causing the wavelength to shorten.
Thanx sir,I am trying to solve this since past 10 days .....
Thnx sir ...❤
Thnx✔
Most welcome
If relativistic objects emit radiation focused in the forward direction ("headlight effect"), then do they reflect radiation the same way?
If so, then if any relativistic object reflects starlight, radar, etc., then it only does so towards its forward direction of travel, and is nearly invisible to the sides or rear?
The first principle that Einstein used in his formulation of the special theory of relativity is that every observer, regardless of how fast the observer is traveling will see light (E&M radiation) approach and move away at the speed of light.
@@MichelvanBiezen Relativistic particles, which radiate isotropically in their own rest frames, appear to radiate preferentially in the forward direction to others?
So, if there was some relativistic object hurtling through our Solar System, reflecting visible-wavelength ambient sun- and star-light, then that object would appear to be bright in gamma rays on the way in, when earth observers were within the "headlight angle" of order 1/gamma. As the object began to pass, it would dim, and the radiation redshift, into x-rays & UV, then dim & redshift further into visible as it passed by at a LOS 90 degrees to its travel. As began to whoosh away, its apparent radiation would dim & redshift further into IR, millimeter, microwave & fade away into the distance as an ultra faint radio source.
To detect such an object would require some sort of multi-band, omni-band, all-band, full-spectrum, gamma-to-radio detector capable of imaging everything from ultra bright to ultra faint.
If a human observer were in orbit, aboard the ISS, unable to see anything but visible wavelengths, they would only sense a brief, fading, visible flash of a streak as the object whooshed past the 90-degree point of closest approach to them. They would conclude cosmic rays were affecting their eyes, and think nothing further of it.
Not sure how you could build an all-spectrum detector, except perhaps to just attach other kinds of detectors (visible, IR, UV) into the gaps in the framework of a large radio dish.
From what I understand right now, something like that would be necessary to recognize the passage of a near-lightspeed object. One would have to overlay all the images from all the different frequency bands in order to recognize that they all "stitched together" into one long streak of a gamma-bright-to-radio-dim track (rather than being a bunch of independent flashes in each of the bands).
I would disagree with your initial statement, since that goes against the principle of the special theory of relativity.
Since T is a period, shouldn't we use the Lorentz transformation T=gamma(T_0 - xu/c^2)
Math Geeks Since we do not care about the coordinates, we can just assume that x in the equation that you have written is 0, and therefore T=gamma•T_0. There is also another simple method to prove the time dilation equation without any coordinate-system. And the result is that T=gamma•T_0.
Hello sir, how did you get from c-u to radical and squared that?it is at 6.49. I didn't get the transition. Did you multiply or how? Would be grateful if you reply and thanks a lot for your videos!
A - B = SQRT { (A - B)^2} = SQRT { (A - B) (A - B) }
excellent video
at what order of the speed of observer should we consider the lorentz transformation? is there any particular speed limit?
Very good question. When v = 10 % of c (the speed of light), then your error will be 0.5 % if you don't use the Lorentz transformation equations. When v = 50 % of c, the error will be 15% and if v = 86.7 % of c then your error will be 100%..
@@MichelvanBiezen ohh ok sir thankyou so much!! btw big fan of your teaching sir!!
I'm not that far along in physics just yet, but does this mean the speed at which I am moving away from the source helps to determine the amount of red shift observed?
Or the amount of the red shift helps to determine your speed or the source speed.
Thanks :)
Thanx for your videos :) I'm more a philosopher than a physician but anyway, let's imagine two cases:
.
1. A star is moving away very fast from an observer on the earth, red-shifted light with a long wave length.
.
2. A star is approaching very fast to an observer on the earth, blue-shifted light with a short wave length.
.
Let's assume that the two stars are now at exactly the same distance from earth.
.
Then let's, theroretically, draw these two waves as two separate straight lines instead (stretch them out). The blue line (former wave) will then be longer than the red one. Is it possible to explain, with this little thought experiment, that the speed of light emanating from the two objects is the same for both of them
(independent of moving away or approaching) ? Basically that photons have to travel a longer distance ("riding on or within" the blue wave), so that the speed in a straight direction (speed of light) stays the same as with the red wave ?
.
Trying to get a grip of this in a real/deeper physical sense if possible, why the speed of light stays the same when objects move differently relative to an observer. Is that a phenomenon which is dependent on some rate of change within the object itself when moving (compared to a rate of change in the observer as an object), or can it be explained occuring within the communication between the objects (my example) ?
Is doppler effect and dopper shift same? Do you have a video for the derivation of
f^'= √((1±β)/(1∓β)) f, where beta = v/c ?
yes, they are the same. The above video is the exact derivation of
f^'= √((1±β)/(1∓β)) f, where beta = v/c
Question sir michel. What if the observer also travels nearly as fast as the source of light, will the Classical Doppler's Effect be used instead of Relativistic Doppler's Effect? Sorry for this question sir.
If the observer travels near the speed of light, all the equations of relativity play a role and yes, the observer will still see light moving at the speed of light regardless of what the source is doing.
@@MichelvanBiezen so how would you relate the frequencies measured by source and observer in that case? t = tO/sqrt(1 + v^2/c^2). tO is applied when the observer isnt moving but it that case, it is moving. Will it be tS = tO/sqrt(1 + vS^2/c^2) and tA = tO/sqrt(1 + vA^/c^2)
Where tS = time measured by source. tA = time measured by observer. t = time measured by stationary object. vS = speed of source moving nearly at c. vA = speed of observer moving nearly at c.
If not, pls make vid about this for clarifications. Thanks sir for your reply.
Very very good.
Thank you! Glad you liked it. 🙂
In the title of this video, should it be "Elected" or "Effect"?
Just asking. Great explanation! keep up the good work :)
It is "effect". Thanks.
Wonderful concept!!Thanks Sir..
Question sir. What if the obsever also moving nearly at speed of light? Will he/she measure "c" (based on 2nd postulate of special relativity) or will he/she measure "c +or- v" where "v" is his/her speed nearly speed of light depending if he/she is away or towards the source?
All observers, regardless of the speed they move at, will observe light to move at the speed of light.
@@MichelvanBiezen thank you sir!!!
Sir please sir🙏🙏 ...humble REQUEST...please help me to know how to derive the time equation which u have taken in the middle🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏
he made a video deriving it
ruclips.net/video/eLGLQ1n6sjs/видео.html
@@tahycoon2.0 thanks bro
@@shaikrafik6300 u welcome
Merci, Michel
C'est mon plaisir :)
What about if the observer is also moving?
well... im 5 months late
if the observing and the source of light are moving towards each other with velocities u and v respectively
then from the reference frame of the observer .. he is at rest but now he will see the light source moving towards him with a velocity of v+u
what if the observer is moving relativisticly too
We have some examples in the playlists where both observers are moving.
Why distance between 1st and 2nd wave is lamda???
lambda is the variable used to express wavelength. (The distance between 2 waves equals the wave length)
Great :) Thanks Dr.
amazing
Is it correct f`=f×(c÷(c-v)) ?
I am not sure what you are asking. Are you asking about the use of the variable "u" instead of the variable "v"?
good videos
Tq re
Nice ... thnxxxx
crank
?
Brooooo wowwwww
Thanx sir,I am trying to solve this since past 10 days .....
Thnx sir ...❤
Thnx✔
Glad it helped!