Question about C and Relativity

  • Thread starter Thread starter CuriousStuden
  • Start date Start date
  • Tags Tags
    Relativity
CuriousStuden
Messages
3
Reaction score
0
So, I was learning about the theory of relativity through Einstein's book "Relativity" (haven't finished calculus yet, so this book works for me), and I had a question about an example.

In the book, Einstein describes the Special Theory of Relativity using the example of a train and the embankment it travels on. He states that, if one were to use classical mechanics and the principle of relativity, a light beam traveling at c relative to the embankment would be c minus the speed of the train, which is proved to be empirically false. To reconcile the two concepts, time is supposed to be dilated so that the beam travels more quickly to a person inside the train.

Now, this is where my question comes in. What if there were two beams, traveling in opposite directions parallel to the train, at c relative to the embankment. To a person on the train, under classical mechanics, one beam would travel at c minus the speed of the train (in the example above) and the other would travel at c plus the speed of the train. Which way is time dilated for the person on the train? If time is elongated, so as to make the beam that travels more slowly than c travel at c, then the other beam travels even faster, and vice versa.
 
Physics news on Phys.org
Time is NOT "dilated" for anyone- except relative to some other frame of reference. In Einstein's example, one frame of reference is on the train, the other on the embankment. In your example, you have given only one frame of reference, the train. The answer to your question is that the person would measure both beams of light as having speed "c".
 
HallsofIvy said:
Time is NOT "dilated" for anyone- except relative to some other frame of reference. In Einstein's example, one frame of reference is on the train, the other on the embankment. In your example, you have given only one frame of reference, the train. The answer to your question is that the person would measure both beams of light as having speed "c".

I only mean "dilated" in the relativistic sense of the word. The example I am trying to say is essentially two of the Einstein examples put together, in opposite directions. Since the beams of light are traveling at "c" relative to the embankment, I am wondering how the time conversion would work when comparing that frame of reference to the train, as a conversion of time either increasing or decreasing from the embankment to the train would run into the principle of the propagation of light.
 
CuriousStuden said:
I only mean "dilated" in the relativistic sense of the word. The example I am trying to say is essentially two of the Einstein examples put together, in opposite directions. Since the beams of light are traveling at "c" relative to the embankment, I am wondering how the time conversion would work when comparing that frame of reference to the train, as a conversion of time either increasing or decreasing from the embankment to the train would run into the principle of the propagation of light.

This involves the "Relativity of Simultaneity". If you are on the train, each light pulse travels at c relative to you. If you are halfway between two points on the train and fire a pulse of light towards each, they will take the same amount of time to reach each point. The pulses will reach the points simultaneously. If there are clocks at these points that are synchronized in the frame of the train, they will read the same time when the light pulses reach them,

If you are on the embankment watching the train go by, the light pulses travel at c relative to you. In this case, one of the points on the train rushes towards the light pulse and the other runs away from it. Thus, in your frame, the arrival of the light pulse at each point are not simultaneous. Since we are talking about the same light pulses, and the same clocks on the train, the clocks still must read the same time when the light pulses reach them. However we have established that according to the embankment, the light pulses do not take the same amount of time to reach each clock. Therefore, according to the emebankment frame, the clocks on the train are not synchronized, even though they are synchronized in the frame of the train.

You also have to take into account that according to the embankment, the distance between the two points of the train are length contracted.

If you have to take all three, time dilation, length contraction and the relativity of simultaneity into account to analyze what happens according to each frame.
 
Oh, ok. Thank you!
 
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Replies
14
Views
1K
Replies
11
Views
1K
Replies
5
Views
2K
Replies
61
Views
5K
Replies
58
Views
6K
Replies
26
Views
1K
Replies
10
Views
2K
Back
Top