The Theoretical Minimum: Length Contraction and Time Dilation

[DavidBalut]

TL;DR Summary

I am trying to understand Minkowski diagrams.

Hello,

My name is Dave and I'm a physics major at UIUC. It looks like I will be taking the special relativity course (phys 225) this fall. I've always been fascinated by the theory so I decided to get a head start with Lenny and Art's perspective on it.

My first head-scratching moment came in section 1.4. First with the "geometry" of length contraction. I completely agree that on pg 38 the line OP is shorter than line OQ, due to the transforming coordinate Q onto the x prime axis in the form of point P. However, what I can't get my mind around is the "geometry of this situation" how can the "leg of a right triangle" be longer than the "hypotenuse of a right triangle"

This happens again on page 42. Time moves slower in the moving frame such that t_stationary frame > t_moving frame. Yet, once again the leg is greater than the hypotenuse.

I have a feeling that apparent lengths on different axes are irrelevant and arbitrary and what really matters is calculated values by Lorentz transformations.

Any help is greatly appreciated!

 

[PeroK]

However, what I can't get my mind around is the "geometry of this situation" how can the "leg of a right triangle" be longer than the "hypotenuse of a right triangle"

Because it's Minkowski and not Euclidean geometry. A line in Minkowski geometry can have zero length if it is light-like. I.e if $\Delta x = c\Delta t$. That's definitely non-Euclidean.

 

[malawi_glenn]

The length L' in S' is defined as Δx' when Δt' = 0

You can use the following for space time diagrams, when S' moves in S in the x-direction with speed v

$\dfrac{\text{one } c t' \text{ unit}}{\text{one } c t \text{ unit }} = \dfrac{\text{one } x' \text{ unit}}{\text{one } x \text{ unit}} = \sqrt{\dfrac{1+ \beta^2}{1-\beta^2}}$ where $\beta = v/c$ it is quite easy to prove this relation with the Lorentz-transformation and it is good to write it down somewhere on a piece of paper and always have it next to you when dealing with space-time diagrams

 

[DavidBalut]

Because it's Minkowski and not Euclidean geometry. A line in Minkowski geometry can have zero length if it is light-like. I.e if $\Delta x = c\Delta t$. That's definitely non-Euclidean.

Oh! That is awesome! Thanks for the quick reply!

So trust the math for values, not the pictures?

 

[malawi_glenn]

So trust the math for values, not the pictures?

The pictures are good, but you need to remember that the units along the ct' and ct axes are not the same. Similarly for x' and x axes, see my post above

 

[DavidBalut]

Aha! I got it now! Thanks so much

 

[malawi_glenn]

Aha! I got it now! Thanks so much

also remember how the axes in the S' system are oriented. You do not measure with right angles

 

[JandeWandelaar]

Summary: I am trying to understand Minkowski diagrams.

I completely agree that on pg 38 the line OP is shorter than line OQ

Isn't OP longer than OQ in the diagram on page 38? In S' you should project OP non-orthogonally on the t'- and x'-axis, parallel to both of them.

 

[Orodruin]

So trust the math for values, not the pictures?

Trust both! But your interpretation of the pictures must be based on the correct geometry.

As already stated in this thread, Minkowski space is not Euclidean space which means you cannot go around applying Euclidean tools such as a graded ruler to go around measuring things.

The difference between the geometries can be traced back to the Minkowski space equivalent of the Pythagorean theorem:
$$
s^2 = t^2 - x^2
$$
(I put $c=1$ because it is convenient)
Note the big difference in the appearance of a minus sign (!) for the spatial part. This means that the Minkowski equvalent of hypothenuse length squared can be positive, zero, or even negative. We refer to such lines as timelike, lightlike (or null), and spacelike, respectively.

You can use this to grade any axes in your Minkowski diagram much in the same way you can use a unit circle to grade axes in Euclidean space. In Euclidean space, you can draw the curve $x^2+y^2=1$. Wherever that crosses an axis (as long as you keep to only ising coordinates with a common origin), that is the 1 mark of the axis.

Similarly, in Minkowski space, the curves $t^2 - x^2 =\pm 1$ are unit hyperbolae. The one with the positive sign will cross any time axis at the 1 mark for that axis and the same for the minus sign but for the spatial axis.

To deduce qualitative results such as length contraction rather than length elongation you do not even need to be very accurate with your hyperbola drawings.

 

[JandeWandelaar]

also remember how the axes in the S' system are oriented. You do not measure with right angles

View attachment 303716

Shouldn't the lower [x',ct'] on the right be [x,ct]?

 

[Orodruin]

Isn't OP longer than OQ in the diagram on page 38? In S' you should project OP non-orthogonally on the t'- and x'-axis, parallel to both of them.

No. OP is shorter than OQ. This is the entire point of length contraction. This is independent of what frame is used to compute it because the length of any spacelike line is invariant.

 

[malawi_glenn]

Shouldn't the lower [x',ct'] on the right be [x,ct]?

it should yes, I just took the first best pic I could find online that showed the "shape" of geometry. I was too tired to scrutinize the labelings

What circle?

I guess he means circle in euclidian space

 

[PeroK]

Oh! That is awesome! Thanks for the quick reply!

So trust the math for values, not the pictures?

Spacetime diagrams, IMO, are not as straightforward as some people like to claim. Especially if you are trying to illustrate a scenario in two different reference frames. As is the case here.

Moreover, if you try to apply your knowledge of Euclidean geometry to them, then they will be misleading.

They can be very useful, nevertheless.

 

[JandeWandelaar]

What circle

I indeed mean the circle in the Euclidean space. The arc-length, which is the same as the angle, corresponds to the arclength of the hyperbola.

 

[Orodruin]

I indeed mean the circle in the Euclidean space. The arc-length, which is the same as the angle, corresponds to the arclength of the hyperbola.

In the correspondence of Euclidean to Minkowski geometry, yes. But with the important point that the arc length is measured with the appropriate geometry, ie, the hyperbola’s arc length measured using the Minkowski geometry. If you draw a hyperbola in a Euclidean plane the argument of cosh and sinh etc are instead related to area. Hence, why the standard names of the inverse hyperbolic functions are arsinh, arcosh etc instead of arcsinh, arccosh, … as many believe.

 

[malawi_glenn]

In order to talk about "look" you need to take into account that ligth has to travel from the event to the observers eyes. Measuring and "seeing" are two separate things in relativity. Look up for instance Terrell rotation of a square. If the square's restlenght is L, the viewer will actually see the length L, but will measure the contracted length L/γ

Lets do the actual calculations (using c = 1 to avoid cluttering) for the space-interval as measured in S': x_P' - x_O'.
In S the event O has coordinates (x_O, t_O) = (0, 0) in S' this event has coordinates (x_O', t_O') = (0, 0)
In S the event P has coordinates (x_P, t_P) = (1, v)
Thus, the space-interval as measured in S is: x_P- x_O = 1
In S' the event P has coordinates (x_P', t_P') = (γ⋅(1-v²), 0) = (1/γ, 0)
Thus, the space-interval as measured in S' is x_P' - x_O' = 1/γ
Since γ ≥1, we have x_P' - x_O' < x_P - x_O
The length in S' as MEASURED in S' is shorter than what it is MEASURED in S.

 

[Sagittarius A-Star]

Events are points yes, but you wrote interval.

No idea what you are after, but the clock in the S frame is stationary in S so it is always positioned there at the (x,t) = (0, t) coordiate. The clock in the S' frame is stationary in S' so its stays positioned at the (x', t') = (0, t') coordinate. Now perform a Lorentz-transformation and you will get that t' = γ⋅(t - 0⋅v) = γ⋅t and this is called time dilation. As you poined out, since the origins of S and S' coincide, we have that Δt' = γ⋅Δt

View attachment 303725

That transforms S-frame-coordinates of a clock $A$, that is at rest in frame S, into coordinates of frame S'.

The 2^nd picture in the OP and the text below show the transformation of the S'-frame-coordinates of a clock $B$, that is at rest in frame S', into coordinates of frame S:
$t_B=\gamma ({t'}_B+0 \cdot v) = \gamma {t'}_B$

I think, that Minkowski diagrams in textbooks should always show scales at their axes for better understanding.

via:
https://www.geogebra.org/m/NnrRvA46

 

[PeterDonis]

@JandeWandelaar you are cluttering up someone else's thread with incorrect information. That is not helpful to either the OP or to other readers. You have now been banned from further posting in this thread.