I took linear algebra during a break once on coursera just to get an idea of it because I was going to start a python/machine learning course. I remember half way through my linear algebra class I remembered a video I casually passed by regarding time dealation a long time ago and instantly drew the connection in how it's calculated. It was interesting for me, as someone who is not math savvy, because I was able to have an "aha!" moment of "so that's how they calculate crazy shi* like that!!"
I am hopeful that you pursue an educating position at some point, maybe after passing physics with flying colors :D great video, and explained really well
Great video! Really clarified the derivation for me. I have a question regarding the measurement of t=1/c. Since time is a speed measured in m/s, woudn't it then be putting time = speed? I can't really grasp the intuition behind such a postulate, hence my comment. Thanks in advance, i love your videos!
@@MuPrimeMath My bad. I meant that the speed of light is measured in m/s. In the start of the video you let t=1/c. The way i understand this is that you are putting time equal to speed, which boggles my intuition. My question is then: How is it possible to let t=1/c, when time is measured in seconds, and c is measured in m/s?
Ok, I see! You're right that taking t=1/c doesn't make sense on its own if we consider the units of t and c. In the context of the derivation, I'm treating all of the variables as just numbers instead of as quantities with units, so I'm just treating t and c as unitless numbers. If you want to keep units for all of the quantities, you could do something like this: "Let z be some quantity in units of distance (e.g. meters), and let t = z/c. Then x = vt = v/c*z, so the vector [x, ct] becomes [v/c*z, z] = z*[v/c, 1]." The rest of the reasoning would be similar to what I did in the video.
γ=1/sqrt(1-(v/c)^2) cos(arctan(v/c))=1/sqrt(1+v^2/c^2) They look very similar, right? What about: cos(arctan(i*v/c))=1/sqrt(1-v^2/c^2)=γ=c/sqrt(c^2-v^2) If cos is used, then what is sin? sin(arctan(i*v/c))=i*v/(c*sqrt(1-v^2/c^2))=i*(v/c)*γ I'm certain I've seen (v/c)*γ being used somewhere. So we have cos and sin. cos(arctan(i*v/c))+sin(arctan(i*v/c)) So, what we have is a complex number, but not yet. In this form we can find the radius by squaring. It is still, technically, a trig function. (r*cos(arctan(i*v/c)))^2+(r*sin(arctan(i*v/c)))^2=r^2 Since it is a complex number, can't we just multiply by it's conjugate? r:=7, v=3*c/4 (7*cos(arctan((i*3*c/4)/c))+7*sin(arctan((i*3*c/4)/c)))(7*cos(arctan((i*3*c/4)/c))-7*sin(arctan((i*3*c/4)/c)))=175 That isn't the right answer is it? Well, the magnitude is (x+i*y)*(x-i*y). We need to multiply by i, even though it's already established to be a complex number. (7*cos(arctan(i*(3/4)))+i*7*sin(arctan(i*(3/4))))(7*cos(arctan(i(3/4)))-i*7*sin(arctan(i(3/4))))=7^2 So, there is this other identity that goes with this. m/m_0=γ The two work together, but I can't work with this as it is. I'm going to get them all on one side. This way the ratio can also be the radius. 1=γ*m_0/m I am starting to see a problem here. If I treat m_0/m like it is the radius, it will be equal to 1. (m_0/m)*γ+i*(m_0/m)*γ*v/c = 1+i*1*v/c= 1+i*v/c. || sqrt(1-(v/c)^2) || which is 1/γ = || m_0/m || 1+i*v/c matches your form. The series of arctan with the Lorentz Factor is also cool. arctan(i*v/c)=i*ln((c+v)/(c-v))/2 where -1 < v/c < 1 Which is actually, arctan(i*y/x)=i*ln((y+x)/(y-x))/2 where -1 < y/x < 1
I always wondered why the LT matrix has determinant 1, now I see it has to do with the symmetry of being able to take the inverse by changing the sign of v
I have a question about the explanation of why the eigenvalues are positive in the description. We can write (β,1) as a linear combination of two eigenvectors. But no matter how they are flipped around due to sign, they remain orthogonal and span the whole space. Thus you can also access the vectors in the region you want as well, so there is no contradiction
The question isn't whether the eigenvectors [1,1] and [-1,1] span the whole space, but rather whether the vector [β,1] will remain in the correct region. If one of the eigenvalues is negative, then [β,1] will not end up in the top region after the transformation, and hence the transformed vector cannot equal [0,1], which violates our assumption. There will be other vectors that end up in the top region after the transformation, but [β,1] will not, and that's the relevant point in this case.
But how is the Mathematics so capable of unentangling a fundamental property of Nature? That I find really perplexing. There is almost no Physics in your presentation but the presentation is about a fundamental property of Physics.
Nonsense. Now redo the math but don't make the light have a 1:1 ratio on your graph. I like to draw light velocity practically along the x axis, because nothing can go faster. This gives me the full plotting area to enter real data. Not only half of it. The weird way you choose to have light at 45 degrees is the only reason you math works the way it does. You have (minkowski has) created a "special case". Only if light is drawn at 45 can you get the result you want to see. The whole theory collapses if you change that one thing.
Thanks for the clear derivation, most satisfying one I've seen
Wow, I've always wondered how and why the Lorentz factor appeared in the time dilation equation, great video!
Finally I found the video I needed.... Thank you my friend for sharing your wisdom and experience with us! You deserve more views and subscribes.
Thanks for putting the time of the actual derivation in the description. Appreciated :)
Nicely done. Thank you for such a refreshing perspective.
Damn this is an incredible explanation, thanks for doing these
Great as usual .
Thank you so much dear *Mu Prime* 💖
I took linear algebra during a break once on coursera just to get an idea of it because I was going to start a python/machine learning course. I remember half way through my linear algebra class I remembered a video I casually passed by regarding time dealation a long time ago and instantly drew the connection in how it's calculated. It was interesting for me, as someone who is not math savvy, because I was able to have an "aha!" moment of "so that's how they calculate crazy shi* like that!!"
7:26 Could this method of eigenvector can be applied for the general case (Lorentz transformation in arbitrary direction) ?
Beautiful derivation!
Loving this SR series
10:54 fun fact, from this point we can notice that the matrix is isomorphic to split-complex numbers, and we can solve the rest in that system.
Great video! Thank you very much!
Great video and it is very clear, thank you for make this content
I am hopeful that you pursue an educating position at some point, maybe after passing physics with flying colors :D great video, and explained really well
Great video! Really clarified the derivation for me. I have a question regarding the measurement of t=1/c. Since time is a speed measured in m/s, woudn't it then be putting time = speed? I can't really grasp the intuition behind such a postulate, hence my comment. Thanks in advance, i love your videos!
I'm not sure what you mean by "the measurement of t=1/c" and "time is a speed measured in m/s".
@@MuPrimeMath My bad. I meant that the speed of light is measured in m/s. In the start of the video you let t=1/c. The way i understand this is that you are putting time equal to speed, which boggles my intuition. My question is then: How is it possible to let t=1/c, when time is measured in seconds, and c is measured in m/s?
Ok, I see! You're right that taking t=1/c doesn't make sense on its own if we consider the units of t and c. In the context of the derivation, I'm treating all of the variables as just numbers instead of as quantities with units, so I'm just treating t and c as unitless numbers. If you want to keep units for all of the quantities, you could do something like this:
"Let z be some quantity in units of distance (e.g. meters), and let t = z/c. Then x = vt = v/c*z, so the vector [x, ct] becomes [v/c*z, z] = z*[v/c, 1]."
The rest of the reasoning would be similar to what I did in the video.
@@MuPrimeMath That makes sense. Thank you so much for taking your time to respond so quickly!
Love from Bangladesh🇧🇩🇧🇩🇧🇩🇧🇩
γ=1/sqrt(1-(v/c)^2)
cos(arctan(v/c))=1/sqrt(1+v^2/c^2)
They look very similar, right? What about:
cos(arctan(i*v/c))=1/sqrt(1-v^2/c^2)=γ=c/sqrt(c^2-v^2)
If cos is used, then what is sin?
sin(arctan(i*v/c))=i*v/(c*sqrt(1-v^2/c^2))=i*(v/c)*γ
I'm certain I've seen (v/c)*γ being used somewhere.
So we have cos and sin.
cos(arctan(i*v/c))+sin(arctan(i*v/c))
So, what we have is a complex number, but not yet. In this form we can find the radius by squaring. It is still, technically, a trig function.
(r*cos(arctan(i*v/c)))^2+(r*sin(arctan(i*v/c)))^2=r^2
Since it is a complex number, can't we just multiply by it's conjugate?
r:=7, v=3*c/4
(7*cos(arctan((i*3*c/4)/c))+7*sin(arctan((i*3*c/4)/c)))(7*cos(arctan((i*3*c/4)/c))-7*sin(arctan((i*3*c/4)/c)))=175
That isn't the right answer is it? Well, the magnitude is (x+i*y)*(x-i*y). We need to multiply by i, even though it's already established to be a complex number.
(7*cos(arctan(i*(3/4)))+i*7*sin(arctan(i*(3/4))))(7*cos(arctan(i(3/4)))-i*7*sin(arctan(i(3/4))))=7^2
So, there is this other identity that goes with this.
m/m_0=γ
The two work together, but I can't work with this as it is. I'm going to get them all on one side. This way the ratio can also be the radius.
1=γ*m_0/m
I am starting to see a problem here. If I treat m_0/m like it is the radius, it will be equal to 1.
(m_0/m)*γ+i*(m_0/m)*γ*v/c = 1+i*1*v/c= 1+i*v/c. || sqrt(1-(v/c)^2) || which is 1/γ = || m_0/m ||
1+i*v/c matches your form.
The series of arctan with the Lorentz Factor is also cool.
arctan(i*v/c)=i*ln((c+v)/(c-v))/2 where -1 < v/c < 1
Which is actually, arctan(i*y/x)=i*ln((y+x)/(y-x))/2 where -1 < y/x < 1
Perfect video
Legal demais. Obrigado
I always wondered why the LT matrix has determinant 1, now I see it has to do with the symmetry of being able to take the inverse by changing the sign of v
It has only been 11 months and I have already forgotten my intuitions on this thing smh
is the fact that the lorentz transform commute with reflections reflects (no pun I swear) the fact that reflections are a symmetry of the universe
What book are you using for this?
which physics and math courses have you taken? and are taking now? If you could do a topology series that would be great.
I probably won't take any topology courses for at least another year, but I'm definitely interested in it!
I have a question about the explanation of why the eigenvalues are positive in the description. We can write (β,1) as a linear combination of two eigenvectors. But no matter how they are flipped around due to sign, they remain orthogonal and span the whole space. Thus you can also access the vectors in the region you want as well, so there is no contradiction
The question isn't whether the eigenvectors [1,1] and [-1,1] span the whole space, but rather whether the vector [β,1] will remain in the correct region. If one of the eigenvalues is negative, then [β,1] will not end up in the top region after the transformation, and hence the transformed vector cannot equal [0,1], which violates our assumption. There will be other vectors that end up in the top region after the transformation, but [β,1] will not, and that's the relevant point in this case.
@@MuPrimeMath got it, thank you!
Is μ=1/γ?
That's correct!
thanks
But how is the Mathematics so capable of unentangling a fundamental property of Nature? That I find really perplexing. There is almost no Physics in your presentation but the presentation is about a fundamental property of Physics.
You get the all the assumptions from Physics
People don't know what a Lorentz transform is and why? Maybe write them out first in equations format
Hope ur jokinh
Thanks! I love the internet.
Nonsense. Now redo the math but don't make the light have a 1:1 ratio on your graph. I like to draw light velocity practically along the x axis, because nothing can go faster. This gives me the full plotting area to enter real data. Not only half of it. The weird way you choose to have light at 45 degrees is the only reason you math works the way it does. You have (minkowski has) created a "special case". Only if light is drawn at 45 can you get the result you want to see. The whole theory collapses if you change that one thing.
You must be very smart
@@alejrandom6592 You just need to be rational and logical. And stay critical and sceptical.
too much gobbledegook
the derivation is much simpler in Ralph Baierlein's Newton to Einstein