SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##

I still wouldn't call it a time dilation factor - that's usually reserved for ##\gamma##. That's not to say it's wrong, just that I've not seen it used that way.I think I see what you're getting at. You're thinking of the classic light-clock in a train moving at constant velocity problem. In that case you do indeed get the time dilation equation because the time it takes for the light to go from one mirror to the other is longer for an observer in the lab than for an observer in the train. Take a look at the time dilation equation - it has the same form as your final equation, except with ##c## replaced by ##v##. But that's because we're dealing with a
  • #1
benorin
Homework Helper
Insights Author
1,435
186
TL;DR Summary
Just a note, wonder where the rabbit hole leads
Suppose the time interval in the Lab frame is a multiple of the time interval in the Rocket frame ##\alpha \Delta t_L = \Delta t_R##, where ##0 < \alpha < 1## without loss of generality. Then the spacetime interval is

##\left( \Delta t_L\right) ^2-\left( \Delta x_L\right) ^2 = \left( \Delta t_R\right) ^2-\left( \Delta x_R\right) ^2\Rightarrow \left( \Delta t_L\right) ^2=\frac{\left( \Delta x_L\right) ^2 -\left( \Delta x_R\right) ^2}{1-\alpha^2}##

The thought ended abruptly there. I guess I just wanted to jot this down before I forgot, and hey, if you've got some good idea how to proceed from here: post it! Thanks!

Edit: something is off, like a minor error... but I can't put my finger on it. Should the bounds for be ##1< \alpha < \infty##?
 
Last edited:
Physics news on Phys.org
  • #2
The spacetime interval between which pair of events? You haven’t specified that, and until you do there’s no way of checking your calculation.

Remember that an event is a point in spacetime, not time. “The moment when the lab clock reads ##T## and the rocket clock reads ##T/\gamma##” is not an event. “The lab clock at this point in space reads ##T##” is an event. “The rocket clock at that point in space reads ##T/\gamma##” is a different event even if it happens at the same time.

So the first step is for you to write down the ##x## and ##t## coordinates of two events. You can use either the lab frame or the rocket frame to assign these coordinates, as long as you’re consistent and only use one frame or the other.

Once you’ve done that and you can calculate the spacetime interval between the two events, use the Lorentz transformations to find the coordinates of the two events in the other frame, and then check that you get the same interval using those coordinates.
 
  • Like
Likes Orodruin
  • #3
As Nugatory says, this depends on the pair of events you are considering. You can make ##\alpha## anything between ##\pm\infty## depending on that choice.

I would guess that you are thinking of ##\alpha## as the time dilation factor, which would imply ##\alpha=\sqrt{1-v^2}## (i.e. your ##\alpha## is the reciprocal of Lorentz' ##\gamma##). If I've guessed correctly, then for your initial equation to hold you need ##\Delta x_R=0##. In that case your final result can be rearranged to read ##(\Delta x_L)^2=(v\Delta t_L)^2##, a restatement of the fact that the rocket moves at ##v## in the lab frame.
 
Last edited:
  • #4
I was indeed as you suspected Ibix thinking along the lines of a rocket going with constant velocity through the Lab frame so that ##\Delta x_R =0 \Rightarrow \alpha = \sqrt{1-\left( \frac{\Delta x_L}{\Delta t_L}\right) ^2}## is what I derive from the OP. Now my question is doesn't ##\frac{\Delta x_L}{\Delta t_L}:=v##? But then I also worked out the moving light-clock problem and suspect this should rather be ##\frac{v}{c}## to give the the expected Lorentz equation (or time dilation equation, whatever?). What am I missing?
 
  • #5
It's usually easier to work in units where ##c=1##, since then you can drop factors of ##c## that pop up all over the place. I did that without commenting - apologies. Your ##\alpha## is ##\sqrt{1-v^2/c^2}## if you don't work in such units, yes.
 

Related to SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##

What is the concept of "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##"?

The concept of "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" is a mathematical expression used in special relativity to describe the distance between two events in spacetime. The ##0<\alpha<1## represents the bounds of the interval, which indicates that the interval is bounded and finite.

How is the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" calculated?

The "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" is calculated using the Minkowski metric, which is a mathematical formula that takes into account the dimensions of time and space. It is calculated by taking the square root of the difference between the squares of the time and space intervals.

What is the significance of the bounds in the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##"?

The bounds in the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" indicate that the interval is finite and cannot be negative. This is important in special relativity because it helps to define the causality of events and ensures that the interval is consistent with the principles of relativity.

How does the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" relate to the concept of spacetime?

The "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" is a fundamental concept in special relativity that helps to define the geometry of spacetime. It is used to measure the distance between two events in spacetime and is essential in understanding the relationship between time and space in the theory of relativity.

Can the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" have a value of zero?

No, the "SR Thought: Spacetime Interval w/ Bounds of ##0<\alpha<1##" cannot have a value of zero. This is because the interval is bounded and cannot be negative, so the smallest possible value is ##\alpha##, which is greater than zero. A value of zero would indicate that the two events are coincident, which is not possible in special relativity.

Similar threads

  • Special and General Relativity
Replies
7
Views
518
  • Special and General Relativity
Replies
1
Views
224
  • Special and General Relativity
4
Replies
123
Views
5K
  • Special and General Relativity
Replies
5
Views
797
  • Introductory Physics Homework Help
Replies
10
Views
937
  • Special and General Relativity
5
Replies
146
Views
6K
  • Special and General Relativity
Replies
20
Views
939
  • Special and General Relativity
Replies
1
Views
1K
  • Special and General Relativity
Replies
1
Views
909
  • Special and General Relativity
Replies
33
Views
2K
Back
Top