B Sun Time vs Earth Time: Lorentz Transformations?

[kent davidge]

When people say that time measured from the Earth is equal to time measured from the Sun minus approx 8 min, are they taking into account Lorentz transformations or simply the fact that light takes approx 8 min from Sun to Earth?

 

[m4r35n357]

When people say that time measured from the Earth is equal to time measured from the Sun minus approx 8 min, are they taking into account Lorentz transformations or simply the fact that light takes approx 8 min from Sun to Earth?

The latter. The Lorentz Transform does not apply to light as it does not have a frame of reference.

 

[PeroK]

When people say that time measured from the Earth is equal to time measured from the Sun minus approx 8 min, are they taking into account Lorentz transformations or simply the fact that light takes approx 8 min from Sun to Earth?

I'm not sure when people would say that and what they would mean by it. The Sun and the Earth are approx 8 light-minutes apart in the rest frame of the Solar system.

 

[kent davidge]

I'm not sure when people would say that and what they would mean by it. The Sun and the Earth are approx 8 light-minutes apart in the rest frame of the Solar system.

Are the times equal?

 

[Nugatory]

Are the times equal?

Which times are you asking about? Using the rest frame of the solar system, a flash of light that leaves the sun at time $T$ reaches the Earth at time $T+8$ minutes.

 

[jbriggs444]

Which times are you asking about? Using the rest frame of the solar system, a flash of light that leaves the sun at time $T$ reaches the Earth at time $T+8$ minutes.

And vice versa. If you shine a flashlight from the Earth at the sun at time $T+8$, the light will reach the sun at time $T+16$. To reasonable accuracy, the situation is symmetric. Neither is "earlier" or "later" than the other.

 

[kent davidge]

Which times are you asking about?

The time measured from a clock on the Sun and the time measured from a clock here on Earth. Wont the NOW in the Sun be different from our NOW?

 

[lomidrevo]

Are the times equal?

It is not very clear what is your question. Are you asking whether two identical clocks (one placed on Earth, second on Sun) tick at the same rate (as seen by an observer on Earth, for example)? The short answer is no.
Btw. the mentioned 8 minutes plays a role in this rate difference (time dilation), but not directly as you probably assume.

 

[jbriggs444]

The time measured from a clock on the Sun and the time measured from a clock here on Earth. Wont the NOW in the Sun be different from our NOW?

If by "now" you mean the hyperplane of simultaneity selected based on an inertial reference frame where the Earth or sun is at rest then, to good accuracy, two such planes the intersect at the Earth "now" or at the Sun "now" will coincide all through the distance between.

 

[PeroK]

The time measured from a clock on the Sun and the time measured from a clock here on Earth. Wont the NOW in the Sun be different from our NOW?

Assuming a clock on the Sun and a clock on the Earth are 8 light-minutes apart in the rest frame of the solar system, then in a reference frame moving at speed $v$ relative to the solar system, the two clocks will be out of sync by $8v/c$ minutes.

In other words, if two events are simultaneous in the rest frame of the solar system, with one event on Earth and the other on the Sun, then those two events can be up to less than 8 minutes earlier or later than each other, depending on your inertial frame of reference.

If you extend the problem to GR, then there are no global inertial frames, so you need a simulaneity convention even to talk about "now" globally.

 

[jbriggs444]

Assuming a clock on the Sun and a clock on the Earth are 8 light-minutes apart in the rest frame of the solar system, then in a reference frame moving at speed $v$ relative to the solar system, the two clocks will be out of sync by $8v/c$ minutes.

Using Earth's orbital velocity for v, that would be about 47 milliseconds -- if the relative velocity were parallel to the 1 AU separation. Since the relative velocity is perpendicular to the separation, the discrepancy in clock setting due to choosing between Sun frame and Earth frame is zero.

 

[lomidrevo]

Using Earth's orbital velocity for v, that would be about 47 milliseconds -- if the relative velocity were parallel to the 1 AU separation. Since the relative velocity is perpendicular to the separation, the discrepancy in clock setting due to choosing between Sun frame and Earth frame is zero.

But ignoring the difference in gravitational potential, correct?

 

[jbriggs444]

But ignoring the difference in gravitational potential, correct?

Right. Of course, gravitational potential (and orbital velocity) affects clock rate, not clock synch per se. The Earth clock may tick more rapidly than the Sun clock, but it will not be "early" or "late".

 

[Nugatory]

The time measured from a clock on the Sun and the time measured from a clock here on Earth. Wont the NOW in the Sun be different from our NOW?

You'll want to understand this in the easier case of flat spacetime (no curvature, no gravitational effects) first.

If we don't worry about gravitational time dilation for a moment:
I can look at my wristwatch to find out what time it is NOW where I am; anywhere else, NOW means "has the same time coordinate as right NOW where I am". Thus, the time on the sun can be pretty much anything depending on our choice of coordinates and especially the simultaneity convention. However, it is natural to assign the time coordinate $T$ to an event if the event happens at a distance $D$ from me and a hypothetical flash of light emitted at that event would reach my eyes at time $T+D/c$; this is the notion of "at the same time" that everyone uses in their daily lives. Using this convention, we would say that whatever happens NOW on the sun we'll be seeing in eight minutes.

 

[PeterDonis]

When people say that time measured from the Earth is equal to time measured from the Sun minus approx 8 min

What people say this, and where? Please give a specific reference. Right now various people are guessing as to what "people" mean when they say this, but without a specific reference, that's all it is, guessing.

 

[PeterDonis]

The Lorentz Transform does not apply to light as it does not have a frame of reference.

This is wrong as you state it. The correct statement is that the Lorentz Transformation acts differently on lightlike vectors than it does on timelike and spacelike vectors. It hyperbolically rotates timelike and spacelike vectors, but it dilates lightlike vectors. You are correct that light does not have a "frame of reference" (more precisely, you cannot construct an inertial frame in which a light ray is at rest), but that does not mean Lorentz Transformations don't act on lightlike vectors.

 

[nitsuj]

You are correct that light does not have a "frame of reference" (more precisely, you cannot construct an inertial frame in which a light ray is at rest), but that does not mean Lorentz Transformations don't act on lightlike vectors.

So does that mean it's because the light is massless is why it cannot have an FOR, not that it can go c.

 

[Ibix]

So does that mean it's because the light is massless is why it cannot have an FOR, not that it can go c.

"Is massless" and "travels at c" are two ways of saying the same thing, because they are both ways of saying "the four momentum is null".

 

[PeterDonis]

So does that mean it's because the light is massless is why it cannot have an FOR, not that it can go c.

"Massless", "cannot have a FOR", and "can go c" are all the same thing, so this question makes no sense.

 

[nitsuj]

"Massless", "cannot have a FOR", and "can go c" are all the same thing, so this question makes no sense.

But light goes c when in a vacuum, not c through stuff, like air or water. what am I missing?

 

[Ibix]

But light goes c when in a vacuum, not c through stuff, like air or water. what am I missing?

The interaction between the light and the matter. You can't really treat light passing through a medium as separate from the medium. And the combination clearly has mass.

 

[Nugatory]

So does that mean it's because the light is massless is why it cannot have an FOR, not that it can go c.

You can argue that position and you won't exactly be wrong, but it's not an especially helpful way of thinking about things. (If all or none of propositions A, B, and C must be true then it is logically sound to claim that A is true because of B not because of C but you aren't gaining any new insight; for that you have to ask why all three are true instead of false).

Better is to start with a postulate that is supported by convincing experimental evidence: the speed of light in vacuum is $c$ in all inertial frames. That trivially implies that there is no inertial frame in which light is at rest; and (with rather more work) gets us to light being massless.

 

[kent davidge]

You say

I can look at my wristwatch to find out what time it is NOW where I am; anywhere else, NOW means "has the same time coordinate as right NOW where I am".

But first you said

You'll want to understand this in the easier case of flat spacetime (no curvature, no gravitational effects) first

Is it because when gravity is important, there's no way to choose a coordinate system such that arbritary points in space have the same time-coordinate? that is (x₁,T) and (x₂,T) is impossible?

Such restrition would imply that x₂ = x₁ + b, with b a constant, correct? Because then dx₁^2 - dT^2 = dx₂^2 - dT^2.

 

[Nugatory]

You say...But first you said...

No contradiction here, the statement about NOW meaning "has the same time coordinate" is always valid, but it easiest to see how it works in flat spacetime (because everything is easier in flat spacetime).

Is it because when gravity is important, there's no way to choose a coordinate system such that arbritary points in space have the same time-coordinate? that is (x₁,T) and (x₂,T) is impossible?

. It's always possible to choose a coordinate system that assigns the same time coordinate to every event on any given space-like hypersurface.