On the way to showing that LT is linear

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In summary, the discussion revolves around the concept of proper time, which is the time interval between two events in the clock frame as measured by the clock. The ratio of d-tau (proper time) and dt (time interval measured in the frame S) should be a constant, regardless of the location and time at which it is measured, as long as the frame S is an inertial frame of reference. This is known as the homogeneity of space and time. However, in non-isotropic situations, such as an accelerated rocketship, this ratio may vary, leading to different rates of ticking for clocks at different heights.
  • #1
neutrino
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It seems to me that words confuse me more than mathematics does. Till the page I'm reading now, Rindler hasn't gone beyond using partial derivatives, and has used more words than mathematics. I don't have anything against it, except that I don't understand what he's trying to say most of the time. Either something is wrong with the way I read the text, or with the way he's trying to explain things. I suspect it is the former. I usually take a break and then reread the parts I don't understand. It helps sometimes, but not in this case.

Here's something from the argument that shows the LT is linear.

Consider a clock C moving with uniform velocity relative to a frame S. The spatial coordinates of the clock in S are [itex]x_i, i = 1,2,3[/itex]. Therefore, [itex]\frac{dx_i}{dt} = const.[/itex]. That's all fine. Here's the part that I don't understand, and I quote from the book.

If [itex]\tau[/itex] is the time indicated by C itself, homogeneity requires the constancy of [itex]\frac{dt}{d\tau}[/itex]. (Equal outcomes here and there, now and later, of the experiment that consists of timing the ticks of a standard clock moving at constant speed.)

Why should [tex]\frac{dt}{d\tau}[/tex] be a constant? I assumed this to be true and read the rest of the argument, which is quite clear.

Thanks for any help.
 
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If d-tau is the time interval between two “ticks” of the clock, as measured by the clock, and dt is the time between these two events as measured in the frame S, then should not this ratio be independent of where and when all this is being measured. S has been assumed to be an inertial frame of ref. That’s what is meant by homogeneity of space and time. So, d-tau/dt should be a constant, i.e., independent of the particular time and the x, y, z co-ordinates of the clock in S. Nothing more, nothing less.
 
  • #3
It might help to think of a case where homogeneity or isotropy is violated for comparison. Suppose you try to apply the arguments to the coordinates (Rindler coordinates) of an accelerated rocketship. Then dt/dtau is not constant, but is a function of height.
 
  • #4
Thanks for the replies.

Shooting star said:
If d-tau is the time interval between two “ticks” of the clock, as measured by the clock, and dt is the time between these two events as measured in the frame S, then should not this ratio be independent of where and when all this is being measured. S has been assumed to be an inertial frame of ref. That’s what is meant by homogeneity of space and time. So, d-tau/dt should be a constant, i.e., independent of the particular time and the x, y, z co-ordinates of the clock in S. Nothing more, nothing less.

That makes sense, sort of. But what does dt/d-tau (as a whole) represent and in which frame is it measured, if it's the ratio of some quantities measured in different frames?

pervect said:
It might help to think of a case where homogeneity or isotropy is violated for comparison. Suppose you try to apply the arguments to the coordinates (Rindler coordinates) of an accelerated rocketship. Then dt/dtau is not constant, but is a function of height.

I'm not yet familiar with Rindler coordinates. I'm trying to do this from first principles, so assume I know only about the postulates of SR and the definition of an inertial frame.
 
  • #5
OK, consider the Harvard tower experiment done on the Earth. Clocks at different heights are sometimes said to tick "at different rates". By this we mean that dt/dtau is different at different heights - dtau/dtau is always trivially equal to one. But this doesn't contradict SR, the situation isn't isotropic, there is a preferred direction, "up".

BTW, technically, I think that this example hows that homogeneity isn't quite enough, one really needs isotropy too. Actually, the original example (based on the "gravitational field" inside an elevator) is required.
 
  • #6
Oh yes, isotropy of course. I should have said "...independent of where and when all this is being measured and in which direction the clock is going."

Let's not discuss the GR effects on clocks here.

Neutrino,

d-tau gives you the "proper time", i.e., the time between two events in the clock frame as measured by the clock, or the interval between two points on the worldline of the clock. It's an invariant. dt will depend on the relative speed between the clock and the IFR S. So, qualitatively, d-tau/dt is a measure of how fast the clock is moving wrt an IFR S. The exact relationship is given by: d-tau/dt=c*sqrt(1-v^2/c^2). Note that when v=0, d-tau=dt.
 
  • #7
Shooting star said:
The exact relationship is given by: d-tau/dt=c*sqrt(1-v^2/c^2).

Sorry for the typo. The 'c' shouldn't be there. The correct formula is: d-tau/dt=sqrt(1-v^2/c^2).
 

Related to On the way to showing that LT is linear

What is the purpose of showing that LT is linear?

The purpose of showing that LT (linear transformation) is linear is to prove that it follows the properties of linearity, which are essential in various mathematical and scientific applications. These properties include preserving addition, preserving scalar multiplication, and preserving the zero vector.

How is linearity demonstrated in LT?

Linearity is demonstrated in LT by showing that for any two vectors u and v in the domain of the transformation, and any scalar c, the transformation of the sum of u and v is equal to the sum of the individual transformations of u and v. Additionally, the transformation of c times u is equal to c times the transformation of u.

Why is it important to prove that LT is linear?

Proving that LT is linear is important because it allows us to use the properties of linearity to simplify calculations and make predictions in various scientific fields such as physics, engineering, and economics. It also helps us to better understand the behavior of systems and processes.

What are some real-life examples of linear transformations?

Some real-life examples of linear transformations include scaling (such as enlarging or shrinking an image), rotation, and reflection. Other examples include converting temperature from Celsius to Fahrenheit, and converting currency from one unit to another.

What are the steps involved in showing that LT is linear?

The steps involved in showing that LT is linear include defining the transformation, stating the properties of linearity, and proving that the transformation follows these properties. This often involves using mathematical equations and properties of vector spaces to demonstrate the preservation of addition, scalar multiplication, and the zero vector.

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