Attention Paid To Accelerating Reference Frames Overthrows SR

[Severian596]

Yes I know this. Let me state for the record:

Clock A is attached to the photon gun, and time measured by it is being denoted by $$\Delta t$$

Clock B was a clock moving at arbitrary speed v, relative to clock A. The Hurkelian result was then derived, where $$\Delta t^\prime$$ is the time of the same event, but measured by clock B.

Up to this point you are good, so what are you trying to do now?

Kind regards,

The Star

Please be careful of your terminology, because Delta t' is NOT an event, it's a measure of change in time between any two defined events. Please don't refer to it as the "time of the same event" unless you've first defined an event (ct,x,y,z) about which to speak. Furthermore the delta t of an event is not a delta at all, it's just t...

Up to this point you are good, so what are you trying to do now?

I'm trying to show you that in my opinion
1) v=c is valid for this equation, and
2) this fact does not contradict special relativity

Why? Well we set up two equations for A's rest frame, right? One solves for Delta t and the other solves for Delta t prime. All with respect to A. If we go ahead and solve each of them for v=c we get the following results:

$$\Delta t = 0$$
$$\Delta t' = Undefined (division by zero)$$

You'll grant me the right to manipulate the equation BEFORE plugging in values, right? So I didn't divide by zero. But why oh why don't these results spell the downfall of Special Relativity??

 

[StarThrower]

Please be careful of your terminology, because Delta t' is NOT an event, it's a measure of change in time between any two defined events. Please don't refer to it as the "time of the same event" unless you've first defined an event (ct,x,y,z) about which to speak. Furthermore the delta t of an event is not a delta at all, it's just t...

I didn't say Delta t` is an event.

Let me define the 'event' I am talking about. The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

Clock B and clock C are in an inertial reference frame, and are at rest with respect to each other. Additionally, they have been synchronized. At the moment in time when clock A and clock B coincide, clock B reads x, and clock C also reads x (because clock C is synchronized with clock B by stipulation). Later, clock A coincides with clock C, and at this moment in time, clock C reads Y. Thus, the total time of the event in the inertial reference frame in which clock B and clock C are at rest, is equal to Y-X. In other words:

$$\Delta t^\prime = Y-X$$

The above quantity is the "time of the event" in reference frame F2.

I have now defined the 'event' I am talking about.

Kind regards,

The Star

P.S. I know why some verbal confusion has arisen.

You refer to a single moment in time as an 'event'. That is the standard terminology yes. But there is another word for this, which is used in philosophical circles, and that word is 'state'.

You represent a state using the following symbolism: (ct,x,y,z)

But you have no precise notion of the "time of an event" which can only be defined using two states, one which occurred before the other, in universal time.

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Regards,

The Star

 

[Severian596]

I thought maybe you'd answer me, or try to tell me that it does in fact lead to a contradiction in SR.

This situation (where v=c) just doesn't make sense. Distance = rate*time. No matter what your rate, if time is zero, distance is zero. But for v=c, no matter what amount of time you try to assert has passed at clock A denoted by $$\Delta t$$, clock B traveling at v=c will measure $$\Delta t = 0$$, which is why we cannot define $$\Delta t'$$ with respect to A. We just wouldn't understand it! No matter how much time we experience traveling with clock A, clock B never measures any change in our clock.

Your initial argument that two photons traveling parallel to each other are traveling relative to each other at velocity zero is classic Newtonian velocity addition. This is an incorrect application of Newtonian physics, just like the application of Newtonian physics to explain the "almost insignificant" change in Mercury's perihelion did not apply (but was precisely explained by Einstein's equations). So, if I can prove that one of your assumptions is incorrect, I can call your entire conclusion incorrect.

I won't mention that you assume that we're in Euclidian space (not proven), and that space is the same for a photon as it is for us (not proven), AND that you assume time is absolute (which I'm sure you know carries implications of variable c and the ether, as well as absolute space, which you've admitted you agree with).

I don't know why I bother. Think what you like, but your thoughts aren't as original as you believe they are.

 

[Severian596]

The event I am talking about begins at the moment...and ends later...

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

 

[Severian596]

I didn't say Delta t` is an event.

Let me define the 'event' I am talking about. The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

Clock B and clock C are in an inertial reference frame, and are at rest with respect to each other. Additionally, they have been synchronized. At the moment in time when clock A and clock B coincide, clock B reads x, and clock C also reads x (because clock C is synchronized with clock B by stipulation). Later, clock A coincides with clock C, and at this moment in time, clock C reads Y. Thus, the total time of the event in the inertial reference frame in which clock B and clock C are at rest, is equal to Y-X. In other words:

$$\Delta t^\prime = Y-X$$

The above quantity is the "time of the event" in reference frame F2.

I have now defined the 'event' I am talking about.

Kind regards,

The Star

Alright, this is convoluted at best, and a diagram may help or it may not, but

The event I am talking about begins at the moment in time when clock A coincides with clock B, and ends later, when clock A coincides with clock C.

I guess I'll assume that, in order to synchronize all three clocks a light source is placed in the center of a Circle (or sphere) in an extremely local spatial area with respect to clocks A, B, and C. A pulse of light is emitted from the light source, and the clocks begin ticking the moment the light reaches them. They are equidistant from the source and not in relative motion to each other or the light source, and therefore we should be confident that they are synchronized. So, in this reference frame F1 (notorious label by now) the lines of simultaneity for clocks A, B, and C are parallel to their spatial axis x with respect to their time axis t.

Clock B and clock C are in an inertial reference frame, and ...

Whoa! When did they change reference frames? You just synchronized them for crying out loud! So we must conclude that now clocks B and C are in a different reference frame F2 than A, then some force must have acted upon them for them. So an acceleration took place. Bad for SR...very bad...you're getting out of special relativity's jurisdiction.

Additionally, they [clocks B and C] have been synchronized.

As long as they've had the same force applied to them and are now in the same reference frame I assume this was not necessary. I'll assume also that you're just reiterating the fact.

At the moment in time when clock A and clock B coincide, clock B reads ...

Okay that's it. Now you're proposing a situation where special relativity does not apply. Because A and B are in different reference frames their lines of simultaneity are NOT parallel to each other. A sees B's lines of simultaneity at an angle, and B sees A's lines of simultaneity at an angle. Therefore there exists no "moment in time" that A and B coincide unless both A and B are in uniform motion with respect to some OTHER reference frame F3. This other reference frame (with respect to which A and B must have been synchronized before hand, then had equal forces applied to each of them to put them in different reference frames than F3 itself) could possibly measure "coinciding" times on A and B, but ONLY RELATIVE TO ITSELF.

Therefore because A and B will never coincide with respect to each other, the rest of your statements never occur.

 

[StarThrower]

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

See my PS which I added to my previous post. It was/is:

P.S. I know why some verbal confusion has arisen.

You refer to a single moment in time as an 'event'. That is the standard terminology yes. But there is another word for this, which is used in philosophical circles, and that word is 'state'.

You represent a state using the following symbolism: (ct,x,y,z)

But you have no precise notion of the "time of an event" which can only be defined using two states, one of which must occur before the other, in universal time.

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Regards,

The Star

 

[Severian596]

Consider two distinct states. Using your notation we have:

State 1 = S1 = (ct1,x1,y1,z1)
State 2 = S2 = (ct2,x2,y2,z2)

Let S1 occur before S2. I am calling [S1,S2] an event, the amount of time of this event is given by your (t2-t1).

Thank you for this. I don't participate in philosophical circles, and we're not in one right now. Your 'event' is actually a duration, I'm sure you know that. Read my previous post to see why your proposed system is invalid...you make assumptions, leave out details, and change frames of reference willy nilly. You know better than that.

Oh and remember that your "Universal Time" you referred to when defining the philosopher's circle 'event' is actually just another frame of reference which is subject to relativity itself. In my post, that "Universal Time" could be in terms of F3, but that means you're not presenting all the necessary facts. And remember that there is no preferred frame of reference, and if there is one (in your opinion) how do YOU personally define that?

 

[StarThrower]

I say again, events don't have a duration. They have 4 coordinates,

(time, x,y,z)

Anyway I'll try to make it through the rest of your post now...though right off the bat you show me that you are in fact confusing the heck out of me by calling an event a duration of time.

*sigh*

I apologize for that, though I knew it would happen. I simply prefer the word 'state' to the word 'event' when I am discussing points of time, or moments in time, and I like to assign quantities of time to something that I wish to call 'events'.

You use the word 'event' to mean what I call 'state', and I use the word 'event' for something you don't have a word for.

Kind regards,

The Star

(P.S. It is easier for you to change than me.)

EDIT: Correction, you do have a word for it... duration. However, duration isn't a noun, the way an event is. Philosophers would have a field day with you.

To familiarize yourself with the philosophical use of the terms 'state' and 'event', peruse the following reputable site which discusses temporal logic.

Temporal Logic

Keep in mind that these guys are just dying to talk with physicists.

 

[Severian596]

duration isn't a noun, the way an event is.

Duration is a noun...and I'm not going to debate about it. Here's dictionary.com's definition (beware an annoying popup).

 

[matt grime]

And linguists would have a field day with you. Not to mention psychologists, given your egoism.

 

[StarThrower]

Duration is a noun...and I'm not going to debate about it. Here's dictionary.com's definition (beware an annoying popup).

Absolutely no point in arguing semantics. I can switch back and forth between your terminology and mine, I am intelligent enough for that. I presume you have the same capability?

Kind regards,

The Star

 

[StarThrower]

And linguists would have a field day with you. Not to mention psychologists, given your egoism.

In the time it takes them to comprehend me, I will have died. Hence, I conclude their efforts would be in vain. Better they should become physicists.

Kind regards,

The Universe

 

[Severian596]

To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

I admire your apparent understanding of physics (both classical and electromagnetic), and your mathematical ability. I'm just a beginning student to the topic(s) yet I spot your error so easily. Why do you persist in thinking you can prove an error in SR where SR does not apply?

 

[StarThrower]

To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

I admire your apparent understanding of physics (both classical and electromagnetic), and your mathematical ability. I'm just a beginning student to the topic(s) yet I spot your error so easily. Why do you persist in thinking you can prove an error in SR where SR does not apply?

Firstly, I didn't make the error someone accused me of, which was to reverse the meanings of $$\Delta t$$ and $$\Delta t^\prime$$

I understand which quantity of time is being measured by which clock. That wasn't/isn't the contradiction I am reaching for. As for your new question about the twin paradox, let me read the article at the site first, and then I will come back and give you my comments. As for my persistence in thinking I have spotted an error in SR... the reason for that is because I have. I don't really believe I am the first to ponder it either. Nonetheless, the error to me seems obvious, yet it hasn't caught on to mainstream physics yet. And that is what I really cannot understand.

If the relative speed in this problem is given by v=c, then we have two photons moving at right angles to each other. The clocks B,C are thus in a photonic frame (see other threads on this). The conclusion I reach, is that if the fundamental postulate of the theory of special relativity is true, then photons cannot move relative to each other. That cannot possibly be true, based upon simple facts that come from sensory perception.

Now let me give this site a look, and I will come back and tell you what I think.

Kind regards,

StarThrower


EDIT: Correction, the clocks B,C are in a photonic reference frame (reference frame in which photons are at rest)

 

[matt grime]

So, you're an expert, a genius able to see to the very inherent contradictory nature of relativity, and you've not come across the twin paradox?

 

[Severian596]

That cannot possibly be true, based upon simple facts that come from sensory perception.

This is the biggest pitfall you could've fallen into. You cannot argue anything based "upon simple facts that come from sensory perception," especially mathematics. Sensory perception is not an axiom. In fact that this phrase even escaped your keyboard surprised the heck out of me. How atrocious.

 

[Severian596]

If the relative speed in this problem is given by v=c, then we have two photons moving at right angles to each other. The clocks A,B are thus in photonic frames (see other threads on this).

Note that this example should be stricken from the record unless StarThrower is willing to better qualify "this problem" and exactly what A and B have to do with any frames, much less "photonic" frames.

 

[StarThrower]

So, you're an expert, a genius able to see to the very inherent contradictory nature of relativity, and you've not come across the twin paradox?

Of course I have, I am merely looking at this site for the first time... then I will give my comments. Immediately upon looking at it I thought to express the population as follows:

$$P(t) = P_0 e^r^t$$

Give me a moment here, I'm reading.

Regards,

Star

 

[Severian596]

Nonetheless, the error to me seems obvious, yet it hasn't caught on to mainstream physics yet. And that is what I really cannot understand.

You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

 

[StarThrower]

This is the biggest pitfall you could've fallen into. You cannot argue anything based "upon simple facts that come from sensory perception," especially mathematics. Sensory perception is not an axiom. In fact that this phrase even escaped your keyboard surprised the heck out of me. How atrocious.

My apologies, allow me to be more professional.

Consider the following experiment performed by an ancient caveman:

He started a fire for the women to sit, directly on his left, so he didnt have to listen to them squalk, and then started a fire directly in front of him, for the men to sit around. He noticed that not only was the front of his body warm, but so was the left side of his body too. He concluded that photons can move relative to each other.

P.S. Or was this a thought experiment? Geeze, I forget. Maybe he just thought he was warm?


Regards,

Star

 

[StarThrower]

You need to understand relativity better. It never caught on because actual physicists (who know more on this topic than I do) evaluated it, pondered it, and dismissed it. Why? Because they know more than you.

They cannot make me believe that which is false. They can confuse me about it for awhile, but they can never make me believe it.

Regards,

StarThrower

 

[Severian596]

He concluded that photons can move relative to each other.

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

 

[Severian596]

And...I shall exit stage left, as I should've done from square one.

Goodbye, StarThrower

 

[StarThrower]

To suppliment my post about your incorrect use of Special Relativity, peruse http://van.hep.uiuc.edu/van/qa/section/New_and_Exciting_Physics/Relativity/20021019150128.htm to the twin paradox. The paradox is not a paradox at all. The individual cites his source and mentions that General Relativity is needed to correctly answer the paradox because Special Relativity does not supply a means to deal with acceleration and/or a preferred frame.

Ok I've looked at the site. The first thing to say, is that I understood Dolan's question.

If you have ever been to the carnival, there is a ride where you are spun around really fast, and then the floor drops out, but you don't fall to the earth. In this ride, you are pinned to the wall by a force which has arisen because you are in uniform circular motion.

Principle of equivalence: It is impossible to tell if you are at rest on the surface of a body exerting a gravitational force F on you, or accelerating uniformly through space.

Dolan asks about the following experiment.

You have two vials of some radioactive substance, with a precisely known half life. For example, suppose that the half life is one hour. The samples are identical, in the sense that they contain the same initial population. So for example, suppose that P0 = 1,000. Thus, if the vials are left at rest, then in one hour, there will be (approximately) 500 radioactive particles left in both samples.

Now, perform this experiment, but this time put one of the samples in a device like the carnival ride described before.

The particles in the centrifuge are obviously subjected to a force which the particles in the vial on the bench are not subjected to.

Dolan's question then, is whether or not the experiment has ever been done, and whether or not any time dilation was observed.

 

[StarThrower]

This line should be changed to read "He concluded that photons can move relative to himself." I can only mildly apprecite your candor.

Imagine that you have two trains A and B that leave the station C like this:

A
^
|
|
|
|
C----------->B

And they just so happen to be traveling at the speed of light. They'll never observe each other, so they'll never determine how fast they're traveling relative to each other.

Umm but they are moving relative to each other, even if they cannot measure the relative speed.

 

[Severian596]

Umm but they are moving relative to each other, even if they cannot measure the relative speed.

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

 

[StarThrower]

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

I thought you exited stage left.

Self-contradiction? Oh now you are totally coming unglued huh. I find a flaw in the theory of SR, and you come apart.

 

[StarThrower]

That's it! That's your flaw. Why should train A or B assume anything based on YOUR measurements. They will never detect each other's presence, so therefore they are NOT moving relative to each other. YOU may determine that they're moving at X speed relative to each other, or Y speed, or whatever speed, but YOU have the luxury of experiencing time and space in a way completely removed from what either of these trains must experience. To them, YOUR claim that they are moving relative to each other is completely preposterous! No matter how long they vigilantly wait for information that let's them draw a conclusion about their motion RELATIVE TO EACH OTHER it will never come.

There IS no absolute preferred frame, and that means YOUR FRAME is not the divine frame from which to ordain all conclusions about the relative motion of any two bodies, or the nature of space and time.

Believe whatever you like. I'm not wasting another moment on you. I've wasted enough.

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

 

[DrMatrix]

moment in Universal time

If you're going to assume SR (or GR), you must drop the notion of Universal time. Otherwise, you will run into a contradiction.

 

[Severian596]

Let them be tied together with a very long string, with a lot of slack in it. Eventually, that string is going to break, and they will 'know' they are/were in relative motion. It will break at the moment in Universal time at which the distance between them exceeds the string length.

Regards,

The Star

Ignoring the impossibility of a string (which has mass) "keeping up" with the photon traveling at c, the compression wave triggered by the snapping string would travel at roughly the speed of sound, just as any compression wave. For example if you and your friend stand with thighs against a table and he bumps it, how much time passes before you feel the table move? The speed is not infinite, and in fact travels at the speed of sound through the medium of the solid table. The photon would never learn of the "universal time" at which the string snapped.

I will discuss the theory of SR with you no further, but am happy to educate you on some basic physical phenomenon.

 

[Hurkyl]

This equation is valid for v=c is it not?

The equation is well-defined, if that's what you mean. In the spirit of the discussion thus far, I'd have to say we currently have no reason to think that the equation would be correct in SR; v=c violates the hypotheses of our derivation.

 

[StarThrower]

The equation is well-defined, if that's what you mean. In the spirit of the discussion thus far, I'd have to say we currently have no reason to think that the equation would be correct in SR; v=c violates the hypotheses of our derivation.

Well, we have the derivation before us, and there was no reason to forbid v=c in the derivation. $$v = \infty$$ certainly must be forbidden, but not v=c. The reason v cannot be infinite, is because as the relative speed v increases, the angle between the sides of the isosceles triangle approaches 180 degrees, at which point the distance D would have to equal zero. But D is nonzero by stipulation. In fact, the only stipulation made about the relative speed, was that it be nonzero (so that we have a triangle in F2).

As we perform algebraic steps, we certainly must be careful not to divide by zero, or WE make an error.

The next thing to say, is that you certainly were correct about my assumption about D, namely that there is no length contraction in a direction perpendicular to the velocity. You also correctly saw that the relative velocity vector was perpendicular to the photon's velocity vector in the lab frame. As for D not contracting, that is unobjectionable as far as I can see, and as for the relative velocity vector, the direction is just one of the stipulations of the experiment.

But, I am sorry to say, v=c is not forbidden by nature, nor was it forbidden in this wonderfully simple derivation of the time dilation formula. Special relativity has an insurmountable problem to contend with, which is namely that photons cannot move relative to one another.

The logic finishes off this way:

Premise 1: If c=c` then (photons can't move relative to one another).
Premise 2: Photons can move relative to one another.
Conclusion: Not (c=c`)

The truth of premise 1 is known deductively, you in fact derived the result for us.

The truth of premise two is known inductively, through our common sensory perception.

The conclusion is sequitur, via the natural deduction known as modus tollens.


Kind regards,

The Star Thrower

 

[Hurkyl]

But, I am sorry to say, there is no reason to forbid v=c in this wonderfully simple derivation of the time dilation formula.

Sure there is.

For instance, you assert:

Now, consider this event as viewed from a reference frame F2 which is moving at speed v relative to F1. In this frame, the path of the photon is not a straight line, but rather the path is triangular

Tell me, how long are the three sides to the triangle? For the sake of argument, let's say D is 1 meter.

There is a second clock (clock B) at rest in F2, and let the time it takes the photon to move from the photon gun to the mirror and back to the photon gun in F2 be measured by clock B to be $\Delta t'$

If, say, $\Delta t$ happened to be 1 nanosecond, what is $\Delta t'$?

In order to synchronize clocks B,C, we can imagine that a rigid ruler connects them, and that we have located the midpoint, and then a spherical light pulse goes off there, and travels to both clocks, setting them to zero simultaneously in reference frame F2. They therefore remain in sync forever in reference frame F2.

How long is the ruler? How long does it take to synchronize the clocks?

 

[StarThrower]

Tell me, how long are the three sides to the triangle? For the sake of argument, let's say D is 1 meter.

The answer comes from Euclidean geometry of course.

We have a right triangle, the hypotenuse of which is S. The base of which is 1/2 v delta t`, and the height of which is D. You want to let D=1 meter.

Additionally, S = 1/2 c` delta t`

Thus

$$(\frac{c^\prime \Delta t^\prime}{2} )^2 = 1 + (\frac{v \Delta t^\prime}{2} )^2$$

Give me a relative speed v in meters per second, and I can finish.

Regards,

Star

 

[Hurkyl]

Give me a relative speed v in meters per second, and I can finish.

You wanted to set v = c; use that.