Confusion about Lorentz Transform

[thrush]

Hi!

I am new here, thought to join as I am trying to learn Relativity, in this case Special Relativity. I have solved a bunch of problems already but ...

The Lorentz Transform formulation I am dealing with is a 4x4 matrix. I understand the invariance of the spacetime interval and have solved it several times:

delta_s^2 = -(delta_t^2) + delta_x^2 ... delta_y, delta_z are zero.

What I don't get is that for any event in frame S, where delta_X (and delta_Y, delta_Z) = 0, and only delta_T is non-zero, I am the guy standing on the train station watching the train pass. Why then is the time in frame S', delta_t' never less than delta_t? It can't be because the denominator sqrt(1-B^2) can never be greater than 1!

How can the passing train's time interval delta_t' be greater than the 'stationary' observer? I thought a velocity like β=3/5C would make delta_t' in the train's inertial reference frame always delta_t' < delta_t?

Sorry to be such a newbie, but I simply must figure this out!

THANKS

 

[Chestermiller]

Welcome to physics forums.

Are you trying to show this from the mathematics of the Lorentz Transformation, or are you trying to understand it on some more fundamental physical basis?

If you are just trying to do it from the mathematics, then there are two ways to do it.

Method 1:

delta_s^2 = -(delta_t'^2) + delta_x'^2 = -(delta_t^2) + 0

Method 2:

If Δx=0, then
$$Δt'=γΔt$$

 

[ghwellsjr]

Hi!

I am new here, thought to join as I am trying to learn Relativity, in this case Special Relativity. I have solved a bunch of problems already but ...

The Lorentz Transform formulation I am dealing with is a 4x4 matrix. I understand the invariance of the spacetime interval and have solved it several times:

delta_s^2 = -(delta_t^2) + delta_x^2 ... delta_y, delta_z are zero.

That's good but it has nothing to do with the following. I presume you already know that.

What I don't get is that for any event in frame S, where delta_X (and delta_Y, delta_Z) = 0, and only delta_T is non-zero, I am the guy standing on the train station watching the train pass.

OK, you've just described your worldline, the sequence of events marking off your Proper Time where you are standing stationary at the spatial origin of frame S. (Note that you haven't described any events for the passing train.)

Why then is the time in frame S', delta_t' never less than delta_t?

Presumably you intend for frame S' to be the rest frame of the train and the coordinates of the events that you have described are how your Proper Time is dilated in the train frame, that is, it takes more Coordinate Time to mark of an equivalent amount of your Proper Time.

It can't be because the denominator sqrt(1-B^2) can never be greater than 1!

Correct.

How can the passing train's time interval delta_t' be greater than the 'stationary' observer?

Because you are only stationary in frame S. In frame S' you are moving and your Proper Time is dilated. Remember, I said you didn't describe any events in frame S corresponding to the moving train. Or you could describe events for the stationary train in frame S' and transform them into frame S and then you would see that they would represent Time Dilation of the moving train.

I thought a velocity like β=3/5C would make delta_t' in the train's inertial reference frame always delta_t' < delta_t?

Yes, the events in your rest frame corresponding to your stationary position and marked off by delta_t are dilated when transformed into the train's rest frame and marked off by delta_t'. Let me add in a couple spacetime diagrams to depict what I have said. The first one is for frame S in which you are stationary at the spatial origin. I have only shown the plot going from 4 microseconds prior to the origin and 4 microseconds after:

Each blue dot represents one microsecond of time in the rest frame S and has coordinates of an event.

Now we transform all the coordinates of the blue dot events into a frame moving at 3/5c towards the right:

Now we see how you are depicted in the train frame: you are moving to the left and your time is dilated or stretched out compared to the Coordinate Time of the train frame.

Sorry to be such a newbie, but I simply must figure this out!

Hopefully you can now.

THANKS

You're welcome.

 

[ghwellsjr]

Now I want to add into the train's rest frame S' events corresponding to it being at the spatial origin of its frame. I'll show the train in red (you are still blue):

I think you can see that the train, at least the engineer's location, in his rest frame looks just like you did in your rest frame.

Now we transform all the events back into the original frame S:

And now we see what I think you originally expected to see of the train's time being dilated.

 

[sardar]

Hi there! It's great that you're interested in learning about Special Relativity and have already solved some problems. The Lorentz Transform can definitely be confusing at first, so don't worry about being a newbie. Let's break down your questions and see if we can clear up some of your confusion.

First, you are correct that the Lorentz Transform is a 4x4 matrix. This is because it deals with both space and time, which are combined in the concept of spacetime. The invariance of the spacetime interval, as you mentioned, is a fundamental principle in Special Relativity and is crucial in understanding the Lorentz Transform.

Now, let's address your question about the time interval in frame S' (the train's frame) being greater than in frame S (the stationary observer's frame). This may seem counterintuitive, but it is a result of the time dilation effect in Special Relativity. Time dilation means that time passes slower for objects that are moving at high speeds. In this case, the train is moving at a high speed relative to the stationary observer, so time on the train appears to pass slower.

The Lorentz Transform takes into account this time dilation effect and adjusts the time interval in frame S' accordingly. This is why delta_t' can be greater than delta_t. It's important to note that this is only true for objects that are moving at high speeds relative to each other, and the difference in time intervals is usually very small.

As for your question about the velocity β=3/5C, this is just an example and the Lorentz Transform can work for any velocity. The key is to remember that the Lorentz Transform takes into account the relative speed between two frames of reference.

I hope this helps clarify things for you. Keep up the good work in your studies of Special Relativity! It's a fascinating subject and can be a bit confusing at first, but with practice and perseverance, you'll get the hang of it. Best of luck!