# Invariant Spacetime Interval

#### kaotak

I am working on a homework problem, and because it is a homework problem, I will not tell you the specifics or ask for an answer. My question has very little to do with the problem in particular, I just wanted to make that disclaimer. Oh, and the derivation of the Lorentz transformation was NOT a homework problem, that was just for fun. I would never post a question about a homework problem without saying it's a homework problem.

Here's the situation of the problem, but not the question: There is an observer on the ground and a particle flies through the atmosphere with constant velocity and then decays.

I checked my answer to the problem by checking that the spacetime intervals (t^2 - x^2) in both the observer's frame and the particle's rest frame are the same. The spacetime interval I refer to is the spacetime interval between the event of the particle's creation and the event of the particle's decay.

I did not, however, get the same spacetime interval for each. Yet I thought that the spacetime interval was supposed to be invariant in all frames of reference. This would mean that the answer I got is wrong.

However, I think I may be mistaken. I looked over my notes and it says that the spacetime interval is invariant under rotation. Perhaps it is not invariant in this situation? For surely, the origins of the observer's frame and the particle's rest frame frames do not coincide. Thus the observer's frame cannot be just a rotation of the particle's frame.

So my question is: should the spacetime intervals in both the observer's frame and the particle's rest frame be the same? And to extend the question to other situations: are the spacetime intervals only invariant under rotation? If the origins of the frames do not coincide, are they still invariant? Is the spacetime interval always invariant, regardless of the setup of any two inertial frames? I would imagine they should be, just thinking about it visually, since intervals don't change length when you shift the graph.

Related Special and General Relativity News on Phys.org

#### jtbell

Mentor
The spacetime interval is invariant both under boosts (velocity transformations) and under spatial rotations. So if you found a different interval in two reference frames, then you made a mistake somewhere in your calculations. Post your calculations in one of the homework forums and someone can probably point out where you went wrong.

#### neutrino

should the spacetime intervals in both the observer's frame and the particle's rest frame be the same?
In SR, spacetime intervals are invariant no matter what. It's similar to distance in Euclidean geometry. Irrespective of how two coordinate systems may be oriented with respect to each other, the distance between two points in space will remain the same.

And to extend the question to other situations: are the spacetime intervals only invariant under rotation? If the origins of the frames do not coincide, are they still invariant?
For the simplest case of standard configuration of two frames, you can easily prove the invariance by plugging in the Lorentz transformations into $x\prime^2 - t\prime^2 = x^2 - t^2$. You can probably use the other, more general equations to prove it for all cases.

Is the spacetime interval always invariant, regardless of the setup of any two inertial frames?
Yes. It wouldn't be such an important aspect of the theory if it weren't.

#### kaotak

Okay, quick question:

Would the x coordinate of the particle always be 0 in its frame of reference? Or would it change? I assumed the x coordinate would always be 0, since in the particle's frame of reference it's still and everything else is moving towards it. But the latter case would make my spacetime intervals invariant.

#### neutrino

It could be zero or some other number, but that number will always be the same in a frame in which the particle is at rest.

#### masudr

kaotak:

The spacetime interval is defined as

$$s^2= \Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2$$

I didn't put the $\Delta$'s in there for fun, but they are usually missing from most definitions, because they are implicit.

If you didn't set up your coordinate systems so that one of them starts off at the origin, then you will need to use the changes in the coordinate variable.

Last edited:

#### pervect

Staff Emeritus
Note that if you use standard units, you need to multiply $\Delta t^2$ by c^2.

#### kaotak

Hmmph. I think I've found my error, and if it's the case, how SILLY!

I did the problem symbolically (as opposed to with numbers) and it comes out right. I believe that the reason why my numbers for the spacetime intervals are different is because my calculator can't store enough decimals to come up with an exact answer. My spacetime intervals are very close to each other in value, and I've repeated the calculations numerous times, so I conclude it's my calculator's fault :P

#### robphy

Homework Helper
Gold Member
So, this sounds like your first lesson in "numerical relativity". To make reliable use of your calculator for this kind of problem, you might have to do a little more algebra and calculus (e.g. Taylor expansion) before plugging in.

#### pmb_phy

kaotak:

The spacetime interval is defined as

$$s^2= \Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2$$

I didn't put the $\Delta$'s in there for fun, but they are usually missing from most definitions, because they are implicit.
I disagree very much with that assumption. The $\Delta$'s means difference in coordinates. When the deltas are absent then it means the difference of one event from the other event which is at the origin, a reference event which is chosen to be the origin of the coordinate system.

If you didn't set up your coordinate systems so that one of them starts off at the origin, then you will need to use the changes in the coordinate variable.

Best wishes

Pete

#### masudr

I disagree very much with that assumption...When the deltas are absent then it means the difference of one event from the other event which is at the origin, a reference event which is chosen to be the origin of the coordinate system.
Erm, what? Did you read the rest of my post? I quote it below

masudr said:
If you didn't set up your coordinate systems so that one of them starts off at the origin, then you will need to use the changes in the coordinate variable.
I'm not sure where the disagreement arises?

Regards,
Masud.

#### kaotak

My interpretation is he agrees with you, and disagrees with the assumption that the deltas are implicit.

#### pmb_phy

Erm, what? Did you read the rest of my post? I quote it below

I'm not sure where the disagreement arises?

Regards,
Masud.
Having dyslexia really sucks sometimes. :(

Sorry Masud. My bad. I humbly appologize for missing that. But with that last part you are 100% right.

Good call and thanks for correcting me.

Pete

ps - A gentleman I know who used to be a physics prof (now retired) suggested that I never start a disagreement with the words "You're wrong." I guess this is one of those places where his wisdom shines through! :)

Best wishes

Pete

#### bernhard.rothenstein

space time interval

I disagree very much with that assumption. The $\Delta$'s means difference in coordinates. When the deltas are absent then it means the difference of one event from the other event which is at the origin, a reference event which is chosen to be the origin of the coordinate system.

If you didn't set up your coordinate systems so that one of them starts off at the origin, then you will need to use the changes in the coordinate variable.

Best wishes

Pete
The problem is I think to establish a relationship between the space time coordinates of the same event. The standard arrangement of the two reference frames makes that in both reference frames we have the events E(0) and E'(0) that take place at their origins at the origin of time.
Presenting the space-time interval as a function of $$\Delta$$ we establish a relationship between distances and time intervals measured in the two frames. If we present it without $$\Delta$$ we establish a relationship between the space-time coordinates of the same event or between the soace coordinates and the readings of the synchrnonized clocks located at the point where the same events take place.
Am I right?

#### masudr

Sorry Masud. My bad. I humbly appologize for missing that. But with that last part you are 100% right.
No worries, mate.

#### gamburch

Creation and Decay

Let's see if I understand the problem. A particle is created and, after some interval of time, decays. The question is: If the particle is moving with respect to an observer, does the observer measure the same decay time as someone moving with the particle. Generally speaking the answer is no!

Information is missing, however, since it matters whether the particle is moving toward or away from the observer, or remaining at the same distance from (as in a cyclotron). In the case of the cyclotron, particles have shown huge apparent differences in half life for observers in the lab. This experiment was done, for example, by Cronin at Princeton sometime in the 50's, I believe.

On the other hand, does an observer in one frame of reference measuring the decay interval of some particle get the same answer as some other observer doing the same in his frame. You bet your sweet bippie he does.

#### bernhard.rothenstein

space-time interval

kaotak:

The spacetime interval is defined as

$$s^2= \Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2$$

I didn't put the $\Delta$'s in there for fun, but they are usually missing from most definitions, because they are implicit.

If you didn't set up your coordinate systems so that one of them starts off at the origin, then you will need to use the changes in the coordinate variable.

Could you give a good reason for the fact that starting with (one space dimensions)
xx-cctt=0
x'x'-cct't'=0
we can consider
xx-cctt=x'x'-cct't'
Thanks

#### gamburch

The basic idea is that a distance between to points is invariant under rotation. The equations you present state the fact for the two space with a signature 0. Another way of looking at the matter is to apply the equations of a Lorentz transformation to x so that you see how x' looks to x. Then substitute these equation into the equation for the line element and you'll get the same line element you had before the Lorentz transformation. That's what it all means to me, and I think others