Discover the Relationship Between E=mc2 and Time with m=1kg and d=1m

[Deepak K Kapur]

If i put c=d/t in
E=mc2, then E=m×d2/t2

Now take m=1kg and d=1m

Does this mean that E is inversely proportional to time?

 

[Fredrik]

If i put c=d/t in
E=mc2, then E=m×d2/t2

Now take m=1kg and d=1m

Does this mean that E is inversely proportional to time?

No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t², that we could say that E is inversely proportional to time squared. But what you found is E= md²/t², where m is a constant and d depends on t. So md² isn't a constant.

 

[PeroK]

If i put c=d/t in
E=mc2, then E=m×d2/t2

Now take m=1kg and d=1m

Does this mean that E is inversely proportional to time?

No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.

Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.

Look up "dimensional analysis".

 

[Deepak K Kapur]

No, that would be the case if there's a constant A such that E=A/t. If there's a constant A such that E=A/t², that we could say that E is inversely proportional to time squared. But what you found is E= md²/t², where m is a constant and d depends on t. So md² isn't a constant.

Okay, i get your point.

But,
1. d=1m and one meter is always one meter. How does 1m depend on time?

2. When d=1m and m=1kg, we get E=1/t2. What is the physical significance of this reduced equation?

Thanks.

 

[Deepak K Kapur]

No, but it means that Energy has a "dimension" of $ML^2T^{-2}$.

Compare this with the classical equation $KE = \frac12 mv^2$, where kinetic energy has the same dimension as above.

Look up "dimensional analysis".

Thanks for your answer.

One more question (silly one).

Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesn't seem to have independent existence.

Then, how on Earth can we multiply a concrete entity with an abstract one??

 

[Fredrik]

1. d=1m and one meter is always one meter. How does 1m depend on time?

1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.

2. When d=1m and m=1kg, we get E=1/t2.

When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$

 

[Dale]

Mass is something concrete, we can see it, touch it, feel it etc.
Speed is something abstract, we can't see it, touch it etc. apart from mass. It doesn't seem to have independent existence.

You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.

Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter into the operation.

 

[Deepak K Kapur]

1 m doesn't depend on time, but this d is defined by d=ct. If it's not, then we're no longer talking about velocity c. So d "depends on" t in the sense that the number represented by the variable d can be calculated if you know the number represented by the variable t.When d=1 m, then t is (1 m)/(299792458 m/s), so what we get is
$$E=(1~\mathrm{kg}) \frac{(1~\mathrm{m})^2}{\left(\frac{1~\mathrm{m}}{(299792458~\mathrm{m/s}}\right)^2} = (1~\mathrm{kg}) (299792458~\mathrm{m/s})^2 =(299792458)^2~\mathrm{kgm^2/s^2} =(299792458)^2~\mathrm{J}.$$

So, it means that in this equation no kind of relation between energy and time can be ever found out...is it so??

 

[Deepak K Kapur]

You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.

Besides, there is no reason that the operation of multiplication requires "abstract" or "concrete" entities to be multiplied by themselves. The concepts of "abstract" and "concrete" don't even enter into the operation.

Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?

 

[Dale]

Mathematical operations are certainly abstract, whether the quantities represented by the math are abstract or not. However, "enters the 'abstract' realm" sounds more like something that you will find in a fantasy novel than in a scientific textbook, so I don't have a direct answer to your question.

 

[Fredrik]

So, it means that in this equation no kind of relation between energy and time can be ever found out...is it so??

Not necessarily. But it means that most candidates for such relations can be immediately discarded because they contradict experiments that don't contradict relativity.

Does it then mean that when we do mathematics ( which is something abstract), then every kind of entity, whether concrete or abstract enters the 'abstract' realm?

I would say that aspects of the real world are represented by abstract mathematical things in the theory.

 

[russ_watters]

2. When d=1m and m=1kg, we get E=1/t2.

No we don't. The units do not disappear.

 

[Deepak K Kapur]

You are confusing mass with matter. You can touch and feel matter. Mass is a property that systems of matter have. It also does not have an independent existence.

So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??

 

[Deepak K Kapur]

No we don't. The units do not disappear.

Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.

Why not, this seems to be a sensible interpretation...

 

[ShayanJ]

Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.

Why not, this seems to be a sensible interpretation...

t can not be a variable. The point is, $c=299792458 \ \frac m s=\frac{299792458 \ m}{1 \ s}=\frac{149896229 \ m}{.5 \ s}=\frac{59958491.6 \ m}{.2 \ s}=\dots$

So you can only have $E=m(\frac{299792458}{1})^2=m(\frac{149896229}{.5})^2=m(\frac{59958491.6}{.2})^2=\dots$.

You're just substituting different representations of the same number c in the formula $E=mc^2$! Its like getting the formula $E_k=\frac 1 2 mv^2$ and writing it as $E_k=\frac{16}{32} mv^{\sqrt{4}}$!

EDIT:
More clearly, if you take d=1 m, then your t is 1/c. So the representation you're using is $E=m(\frac{1}{\frac{1}{c}})^2=m(\frac{1}{\frac{1}{299792458}})^2 \approx m(\frac{1}{3.33 \times 10^{-9}})^2$!

 

[russ_watters]

Cant we say in simple language that when a mass of 1kg moves a distance of 1m then as per E=mc2, we have E=1/t2.

Why not, this seems to be a sensible interpretation...

Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.

So, can it be said that when we do science or math, we are not concerned with the 'real' things but are only bothered about their properties??

Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.

 

[Deepak K Kapur]

Well, two reasons:
1. Because the units are still there and you are pretending they aren't. The "1" has units kg-m2. The equation should read E=1 kg-m2/t2.
2. The equation isn't describing the motion of a kilogram of mass, the piece you broke apart is describing the speed of light ("c" is the speed of light). So the 1m is the distance traveled by light in "t" time: time is a constant here, not a variable and you can't change that. So when you plug in m=1 kg, and d= 1m, t=1/300,000,000s and it all simplifies to E=9x10^16 Joules.

Science includes the implied assumption that the observed properties are real and there is nothing else to "concern with" or not "concern with". Beyond that, you're getting into philosophy, not science.

I don't get your point fully...

1. 'c' in this equation implies that motion of mass is involved. So, why not describe the motion of a kg of mass?

2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?

 

[ArmanCham]

Time constant means this ;
c=x/t If we assume x is one then you cannot change t cause c is constant.So you are telling us E=1/t² here you assumed x equal one.Then you assumed t =2 We cannot change time If you choose x =1.If you change time you can see easily that you are breaking fundamental physics law "Speed of light is constant"
Here the math c=x/t you said x =1 then t=1/c.Now we found t.Its a number but you are telling it can be three or four.Thats nonsense.

 

[russ_watters]

1. 'c' in this equation implies that motion of mass is involved.

You are mistaken. C is just a universal constant, that happens to be the speed of light.

2. How is time constant here? If i put t= 2 sec, the distance gets doubled to keep the speed of light constant. So, how is time constant?

You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.

You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.

 

[Deepak K Kapur]

You are mistaken. C is just a universal constant, that happens to be the speed of light.

You stated the distance was 1m, which just makes time the new constant in the equation. No matter how you change the distance, the time must change in the opposite way in order that d/t always equals C. That's why plugging-in C=d/t is just an unnecessary complication: C is always going to be the same, so there is no point in calculating it every time you do the problem.

You really need to stop trying to break the equation. If you get to a point where you think you have, all it means is that you've confused yourself.

I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

2. What does c in this equation mean?

Thanks, u hav been very helpful...

 

[russ_watters]

1. if v in KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

There is no "implies" here: V is the speed of the object, C is the speed of light. Though the two equations bear some similarities, they are not the same equation.

2. What does c in this equation mean?

Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.

http://en.m.wikipedia.org/wiki/Mass–energy_equivalence

 

[Fredrik]

1. if KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.

 

[Deepak K Kapur]

Again, c is the speed of light. Or do you really mean what does the equation mean? The equation provides a conversion that demonstrates the equivalence of matter and energy. So, for example, if you weigh the fuel of a nuclear reactor after it is spent and find it to be less than its starting weight, this equation tells you how much energy was released by converting the lost matter into energy.

http://en.m.wikipedia.org/wiki/Mass–energy_equivalence

Actually i was expecting the meaning of this equation as follows..

Matter will change into energy when......(some relation to the speed of light)
Cant you elaborate this equation in this way?

Also please explain the meaning of 'square' of c?

 

[Deepak K Kapur]

In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.

Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?

If they are same, then do they look different to the 'observer' only or are they 'really' different?

 

[russ_watters]

Actually i was expecting the meaning of this equation as follows..

Matter will change into energy when......(some relation to the speed of light)
Cant you elaborate this equation in this way?

No, that isn't what it means. I can't really elaborate on something that isn't true.

Also please explain the meaning of 'square' of c?

C^2 is the conversion factor to equate matter and energy.

 

[russ_watters]

Also, i read on the net that one can have c=1 also. If this is the case, what meaning is left in 'conversion' then. If c=1, does it mean that energy and mass are same?

If they are same, then do they look different to the 'observer' only or are they 'really' different?

Not the same, equivalent.

Note, the actual value of C depends on what units you express it in. It can be anything as long as those units are consistent throughout the equation.

 

[Deepak K Kapur]

Not the same, equivalent.

Would you like to elaborate?

 

[russ_watters]

Would you like to elaborate?

Equivalent in value, but not the same in subtance (if that word even applies). IE, two apples vs two oranges.

Please read the wiki link I provided.

 

[Deepak K Kapur]

I will not argue further on the time issue...
But,
1. if v in KE=1/2mv2 implies the motion of mass, why doesn't c in E=mc2 imply the motion of mass.

2. What does c in this equation mean?

Thanks, u hav been very helpful...

Somehow, i can't let it go...please bear with me

1. In KE=1/2m × v2, if we replace the multiplication sign '×' by the addition sign '+' (just suppose), we get, KE=1/2m +v2.

In other words '×' means that the object'moves' with velocity 'v'
whereas
'+' does not mean so.

2. So, why does '×' in E=m x c2 does not mean that the object moves with c (or near c), even if 'c' is the speed of light. It's speed afterall...

 

[jbriggs444]

1. In KE=1/2m × v2, if we replace the multiplication sign '×' by the addition sign '+' (just suppose), we get, KE=1/2m +v2.

This replaces an equation that is motivated by centuries of experiment and replaces it with something that was conjured up with the stroke of a pen and which is not even dimensionally consistent.

In other words '×' means that the object'moves' with velocity 'v'
whereas
'+' does not mean so.

That is not what times and plus mean.

 

[Vanadium 50]

Somehow, i can't let it go...please bear with me

First, enough with the text speak. It's against PF Rulez...er...Rules. You've been warned before, and the fact that you continue to do this shows a disrepect for the forum membership - the same people you are asking for help. You can do better than this.

Second, you can't just start with an advanced topic like relativity until you understand the foundations it is built on. You clearly have gaps in your understanding going back to at least Newtonian mechanics and possibly arithmetic. You'll need to fill those gaps before you can understand the ideas that build upon them.

 

[russ_watters]

Somehow, i can't let it go...please bear with me

1. In KE=1/2m × v2, if we replace the multiplication sign '×' by the addition sign '+' (just suppose), we get, KE=1/2m +v2.

In other words '×' means that the object'moves' with velocity 'v'
whereas
'+' does not mean so.

I'm sorry, but since the units don't match, that's just mathematical gibberish. I don't mean to sound insulting here, but how much schooling have you had? What is the highest level of math and science you've completed? In the US, we typically start with a course at age 14 that teaches the basic tools of science, including dimensional analysis. I've you've passed that level, you'll need to go back and brush-up on it. Here's a link discussing it:
http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html

2. So, why does '×' in E=m x c2 does not mean that the object moves with c (or near c), even if 'c' is the speed of light. It's speed afterall...

Because it was derived to mean something else. Did you read the wiki link I posted? I guess it's unfortunate that the proportionality constant happens to be C (squared), but that's all it is in this equation: a proportinality constant. C happens to be the speed of light, the maximum speed of objects (asymptotic) and a proportionality constant relating mass and energy. It can and does served different purposes in different equations. You're not alone here, it does cause a lot of confusion. IE:

Einstein's famous equation tells us that mass moving at near light speed is essentially energy.

I'm sorry, but no, it really doesn't. As already said by someone else, e=mc² is valid only for stationary objects.

It's not exactly the same, but you might think of it as being similar to how both torque and work have the same units but mean two very different things.

 

[russ_watters]

So I googled the question for other ways to say the same thing and found this relevant quote from a 13 year old PF thread:

...we know, from relativity that energy and mass are proportional. That is, Energy is some constant times mass. We know, by comparing units (dimensional analysis) that the constant must have units of "speed". And the only fixed speed in the universe is the speed of light. That's very much a "hand waving" explanation but a more accurate explanation would have to be how the equation was derived in the first place and a link to that has already been given.

And:

If you have some proportionality between energy and mass, E=bm, then just by looking at the units you know that b needs to have units of J/kg=m²/s². The only combination of fundamental constants that has those units is c², so it has to be c².

Of course, that begs the question, why is it 1 c² instead of 5 c²? And why is it E=bm instead of E=bm²? To answer either of those you really need a full derivation.

 

[DarioC]

Hope I don't regret this: Lot of complex answers here so far.

C= does not equal distance over time except for the values that are measured when determining the speed of light. C is a specific constant in this formula and cannot be used as a variable as you are trying to do because it is not true for any other values.

E=MC squared says that if you convert a certain mass to energy the amount of energy will be the mass times the speed of light squared, not d over t squared, except for the one ratio that gives the speed of light. A little bit of mass, a lot of energy.

The formula C=d/t is a specific version of V=d/t and only has one value for C, even though it can be expressed in different units that will give different units of E. (as has been said already).

You, or I, can do all kinds of things with mathematics. Sometimes the results are true and sometimes they are way wrong.

DC

 

[Deepak K Kapur]

First, enough with the text speak. It's against PF Rulez...er...Rules. You've been warned before, and the fact that you continue to do this shows a disrepect for the forum membership - the same people you are asking for help. You can do better than this.

Second, you can't just start with an advanced topic like relativity until you understand the foundations it is built on. You clearly have gaps in your understanding going back to at least Newtonian mechanics and possibly arithmetic. You'll need to fill those gaps before you can understand the ideas that build upon them.

Well...i was expecting such a response...

I am a teacher myself but never snub anyone who tries to understand something honestly, even if the quality of questions posed is dismal...

I also don't threat such a person...

Well, some powerful people tend to do such a thing...

 

[russ_watters]

Well...i was expecting such a response...

I am a teacher myself but never snub anyone who tries to understand something honestly, even if the quality of questions posed is dismal...

I also don't threat such a person...

To be perfectly frank, it doesn't seem like you are trying very hard here. I agree with V50: you can do better. Especially if you are a teacher.

 

[Deepak K Kapur]

I'm sorry, but since the units don't match, that's just mathematical gibberish. I don't mean to sound insulting here, but how much schooling have you had? What is the highest level of math and science you've completed? In the US, we typically start with a course at age 14 that teaches the basic tools of science, including dimensional analysis. I've you've passed that level, you'll need to go back and brush-up on it. Here's a link discussing it:
http://www.chem.tamu.edu/class/fyp/mathrev/mr-da.html

Because it was derived to mean something else. Did you read the wiki link I posted? I guess it's unfortunate that the proportionality constant happens to be C (squared), but that's all it is in this equation: a proportinality constant. C happens to be the speed of light,

I know that + sign is wrong, i introduced it to show that when we have a mass 'm and a velocity 'v' multiplied together, it means that 'm' moves with 'v'.

Whereas, m+v does not show movement of mass..

Can you give another example of an equation (other than E=mc2), that has mass multiplied by velocity/speed but does not involve motion of mass?

 

[Deepak K Kapur]

To be perfectly frank, it doesn't seem like you are trying very hard here. I agree with V50: you can do better. Especially if you are a teacher.

I am trying very hard..

Well, i am not a science teacher...

 

[DarioC]

Ah I think I get the concept problem that you are having. The C in the formula is just a short way of saying that the calculation of the conversion of matter to energy in a non-moving mass (like in a nuclear reactor ) requires the use of a constant that is equal to the speed of light (because of it's derivation). The formula has nothing to do with velocity or a moving object-- as used.

Perhaps if you could think of the letter C in this formula as just being a number (C stands for constant, chuckle) that happens to be the same as the speed of light (because of it's derivation) you would stop thinking of it meaning that the mass object in the formula must be moving.

Sorry about he liberties I'm taking here guys, I use to write tech manuals and some of them had to get very basic. Old habit.

DC

 

[Deepak K Kapur]

Ah I think I get the concept problem that you are having. The C in the formula is just a short way of saying that the calculation of the conversion of matter to energy in a non-moving mass (like in a nuclear reactor ) requires the use of a constant that is equal to the speed of light (because of it's derivation). The formula has nothing to do with velocity or a moving object-- as used.

Perhaps if you could think of the letter C in this formula as just being a number (C stands for constant, chuckle) that happens to be the same as the speed of light (because of it's derivation) you would stop thinking of it meaning that the mass object in the formula must be moving.

Sorry about he liberties I'm taking here guys, I use to write tech manuals and some of them had to get very basic. Old habit.

DC

Not just a chuckle, u can have a hearty laugh...

But for me the math involved is very strange ( or u may take it as a proof of my foolishness)...

Two same quantities [mass and speed ( whether constant or not)] are involved in the same mathematical relation of multiplication...and... in one case the operation of multiplication says that the object moves and in the second it says it doesn't...u may feel okay with it but i can't.

Mathematics represents nature...what kind of weird representation is this...i wonder.

 

[ShayanJ]

Not just a chuckle, u can have a hearty laugh...

But for me the math involved is very strange ( or u may take it as a proof of my foolishness)...

Two same quantities [mass and speed ( whether constant or not)] are involved in the same mathematical relation of multiplication...and... in one case the operation of multiplication says that the object moves and in the second it says it doesn't...u may feel okay with it but i can't.

Mathematics represents nature...what kind of weird representation is this...i wonder.

Now that it got to this point, I can only have one advice for you. Put aside this question and start learning physics the right way. Then after learning a bit of special relativity, you'll find out that why you are wrong and what is the right way of thinking about this. Keep in mind that you can't expect to have a good understanding of something if you're not willing to put enough effort in learning it.

 

[DarioC]

Uug, too early in the AM for this stuff.

I just pulled out my book on relativity by Albert himself, published in 1931 and from a little quick review I think that you may well have a good point with your fixation (chuckle) with the velocity of light part of the formula. I refreshed my brain that the mc squared is actually a part of the calculations involving the energy and equivalent mass of light when dealing with different inertial frames. As in: I'm moving relative to you.

Strangely enough I myself have never seen in all my books anything that I identify as a connection between those formulas and the energy given off by a nuclear reaction. The only thing that occurs to me is that most of the energy given off by an atomic bomb is initially light. So your moving/non-moving point is better taken by me now.

As the gentleman said above, you are going to have to dig in on the basics. You will really like what Poincare (sp?) and Lorentz have to say.

Good luck
DC

 

[Deepak K Kapur]

There

Uug, too early in the AM for this stuff.

I just pulled out my book on relativity by Albert himself, published in 1931 and from a little quick review I think that you may well have a good point with your fixation (chuckle) with the velocity of light part of the formula. I refreshed my brain that the mc squared is actually a part of the calculations involving the energy and equivalent mass of light when dealing with different inertial frames. As in: I'm moving relative to you.

Strangely enough I myself have never seen in all my books anything that I identify as a connection between those formulas and the energy given off by a nuclear reaction. The only thing that occurs to me is that most of the energy given off by an atomic bomb is initially light. So your moving/non-moving point is better taken by me now.

As the gentleman said above, you are going to have to dig in on the basics. You will really like what Poincare (sp?) and Lorentz have to say.

Good luck
DC

There is a popular ghazal ( a type of song) in India. It has a couplet...i translate it below...

'' Now that i have turned unconscious, people have come to console me''

 

[DarioC]

Well I had read through Einstein's book in bits and pieces over the years, but you got me going again this morning and I spent about two hours, starting at the beginning and reading properly. So thanks for getting me curious again.

Check back on this for a while, I am following up on what I said above and I think I will have more to say on the subject later, after cross-checking my initial findings.

It is very interesting.
DC

 

[russ_watters]

Can you give another example of an equation (other than E=mc2), that has mass multiplied by velocity/speed but does not involve motion of mass?

So, I don't know of another one with speed, but for other equations with similar issues, I gave you an example of torque vs work, that you didn't appear to notice. For that matter, then, any equation with distance in it (circumference or area of a circle, area of a square, etc.) is different than torque and work even though all of them include distance. There is also the equation for weight (and hydrostatic pressure), which uses gravitational acceleration even though the object isn't accelerating (there is a good reason for that, though).

 

[Dale]

Two same quantities [mass and speed ( whether constant or not)] are involved in the same mathematical relation of multiplication...and... in one case the operation of multiplication says that the object moves and in the second it says it doesn't...u may feel okay with it but i can't.

You shouldn't feel OK with it. Your argument is so wrong that you cannot help but feel confused.

First, they are not they are not "two same quantities" they are "two pairs of quantities with the same units". PeterDonis has already pointed out several other examples.

Second, the operation of multiplication NEVER "says that the object moves". The operation of multiplication is simply a mathematical operation, not a physics statement. If I buy n=4 cups of coffee for p=$2/cup then the cost is k=np. That in no way means that either the cups or the price is moving.

Also, consider the fact that multiplication is commutative. So if the multiplication in $mv^2$ meant m moves with speed v then by the commutative property you would have to also say $v^2$ moves with speed $\sqrt{m}$. Hopefully that illustrates how absurd this argument is.

In order to use a formula in physics you need to know the definition of each variable and further you need to know when it applies. You cannot randomly change the meaning of variables and expect it to still work, nor can you apply it when it doesn't apply.

In $E=1/2\,mv^2$ where m is the mass of an object, v is its speed, and KE is its kinetic energy. It only applies when the object has a uniform velocity (not rotating or deforming). The fact that the object moves at speed v is due to the definition of v and not the operation of multiplication.

In $E=mc^2$ m is the mass of the object, c is the speed of light, and E is the total energy. This equation applies only when the object has no net momentum. In other words, $mc^2$ is the energy at rest (v=0), it is not the energy of an object moving at c.

 

[russ_watters]

I think the argument was 10% less bad than that, dale: he wasn't saying that when multiplied by anything it represented motion, only when multiplied by speed (or distance?). Not quite as wrong, but still pretty wrong. That's why I like the example of work vs torque: exactly the same units but one involves motion and the other doesn't.

 

[Deepak K Kapur]

I think the argument was 10% less bad than that, dale: he wasn't saying that when multiplied by anything it represented motion, only when multiplied by speed (or distance?). Not quite as wrong, but still pretty wrong. That's why I like the example of work vs torque: exactly the same units but one involves motion and the other doesn't.

Although i have been warned by ADMIN, let's hope he takes a lenient view...

Let me start from the beginning...

KE=1/2mv2 was empirically found by firing projectiles into water.

It was found
1. KE is proportinal to mass of the projectile.
2. KE is proportinal to the square of speed of the projectile.

So they/he wrote KE=mv2 (1/2, the proportionality constant was added later on to give the equation a clean look, because previously the constant was somewhat cumbersome---you must be knowing all this)

But...my question is very silly rather utterly insane...

Why did he write mxv2 and not m+v2?
I know u may say this is common sense or that only multiplication depicts reality here.

But again...
Why does only multiplication express the reality here?

 

[russ_watters]

KE=1/2mv2 was empirically found by firing projectiles into water...

...1/2, the proportionality constant was added later on to give the equation a clean look, because previously the constant was somewhat cumbersome---you must be knowing all this)

Near as I can tell, neither of those is true. Where did you get them? The wiki does not include them and today, kinetic energy is derived mathematically.

Why did he write mxv2 and not m+v2?

Because that's what "proportional to" means.

For example, in a direct, linear proportion, doubling the independent variable doubles the dependent variable. In a square proportion, doubling the independent variable increases the dependent variable by a factor of four.

If you have two dependent variables, it should be easy to see that only multiplication enables both to act according to the definition of "proportion".

All that said, this has very little to do with your problem, unless this is the entry point into the next (first?) four years of your math education beyond arithmetic.

 

[russ_watters]

Hello DC
Whoever told you that Einsteins equation applies only to stationary objects doesn't really understand the meaning of it. After all c is the speed of light so the equation is only useful when things are moving very fast.

You are completely wrong. Please stop saying this and read the wiki article on mass-energy equivalence I provided. The words "rest mass" or energy literally appear in it a hundred times.

It was Frederick who fist said it:

In non-relativistic classical mechanics, the total energy of a free particle of mass m moving at speed v is $\frac 1 2 mv^2$. In special relativity, it's $\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$. Note that this is equal to $mc^2$ if and only if $v=0$.

And note that since both c and v appear in the equation, it should be obvious that they are totally different things. One is the speed of the object and the other is the speed limit of the universe.